2.2: Derivatives: Properties and Formulas
- Page ID
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)This section begins with a look at which functions have derivatives. Then we’ll examine how to calculate derivatives of elementary combinations of basic functions. By knowing the derivatives of some basic functions and just a few differentiation patterns, you will be able to calculate the derivatives of a tremendous variety of functions. This section contains most — but not quite all — of the general differentiation patterns you will ever need.
Which Functions Have Derivatives?
A function must be continuous in order to be differentiable.
If: a function is differentiable at a point
then: it is continuous at that point.
It is vital to understand what this theorem tells us and what it does not tell us: If a function is differentiable at a point, then the function is automatically continuous there. If the function is continuous at a point, then the function may or may not be differentiable there.
Proof
Assume that the hypothesis (\(f\) is differentiable at the point \(c\)) is true. Then \(\displaystyle \lim_{h\to 0} \, \frac{f(c + h) - f(c)}{h}\) must exist and be equal to \(f'(c)\). We want to show that \(f\) must necessarily be continuous at \(c\), so we need to show that \(\displaystyle \lim_{h\to 0} \, f(c + h) = f(c)\).
It’s not yet obvious why we want to do so, but we can write:\[f(c + h) = f(c)+ \frac{f(c + h) - f(c)}{h} \cdot h\nonumber\]and then compute the limit of both sides of this expression:
\begin{align*}
\lim_{h\to 0} \, f(c + h) &= \lim_{h\to 0} \, \left(f(c)+ \frac{f(c + h) - f(c)}{h} \cdot h\right)\\
&= \lim_{h\to 0} \, f(c)+ \lim_{h\to 0} \, \left(\frac{f(c + h) - f(c)}{h} \cdot h\right)\\
&= \lim_{h\to 0} \, f(c)+ \lim_{h\to 0} \, \left(\frac{f(c + h) - f(c)}{h}\right) \cdot \lim_{h\to 0}\, h\\
&= f(c) + f'(c) \cdot 0 = f(c)
\end{align*}
Therefore \(f\) is continuous at \(c\).
We often use the contrapositive form of this theorem, which tells us about some functions that do not have derivatives.
If: \(f\) is not continuous at a point
then: \(f\) is not differentiable at that point.
Show that \(f(x) = \left\lfloor x \right\rfloor\) is not continuous and not differentiable at \(x=2\).
Solution
The one-sided limits \(\displaystyle \lim_{x\to 2^{+}}\, \left\lfloor x \right\rfloor = 2\) and \(\displaystyle \lim_{x\to 2^{-}}\, \left\lfloor x \right\rfloor = 1\) have different values, so \(\displaystyle \lim_{x\to 2}\, \left\lfloor x \right\rfloor\) does not exist. Therefore \(f(x) = \left\lfloor x \right\rfloor\) is not continuous at \(2\), and as a result it is not differentiable at \(2\).
Lack of continuity implies lack of differentiability, but the next examples show that continuity is not enough to guarantee differentiability.
Show that \(f(x) = \left| x \right|\) is continuous but not differentiable at \(x = 0\).
Solution
We know that \(\displaystyle \lim_{x\to 0}\, \left| x \right| = 0 = \left| 0 \right|\), so \(f\) is continuous at \(0\), but in Section 2.1 we saw that \(\left| x \right|\) was not differentiable at \(x = 0\).
A function is not differentiable at a cusp or a “corner.”
Show that \(f(x) = \sqrt[3]{x} = x^{\frac13}\) is continuous but not differentiable at \(x = 0\).
Solution
We can verify that \(\displaystyle \lim_{x\to 0^{+}} \, \sqrt[3]{x} = \lim_{x\to 0^{-}} \, \sqrt[3]{x} = 0\), so \(\displaystyle \lim_{x\to 0} \, \sqrt[3]{x} = 0 = \sqrt[3]{0}\) so \(f\) is continuous at \(0\). But \(\displaystyle f'(x) = \frac13 x^{-\frac23} = \frac{1}{3\sqrt[3]{x^2}}\), which is undefined at \(x = 0\), so \(f\) is not differentiable at \(0\).
A function is not differentiable where its tangent line is vertical.
At which integer values of \(x\) is the graph of \(f\) in the figure below continuous? Differentiable?
- Answer
-
\(f\) is continuous at \(x = -1\), \(0\), \(2\), \(4\), \(6\) and \(7\). \(f\) is differentiable at \(x = -1\), \(2\), \(4\), and \(7\).
Graphically, a function is continuous if and only if its graph is “connected” and does not have any holes or breaks. Graphically, a function is differentiable if and only if it is continuous and its graph is “smooth” with no corners or vertical tangent lines.
Derivatives of Elementary Combinations of Functions
We now begin to compute derivatives of more complicated functions built from combinations of simpler functions.
The derivative of \(f(x) = x\) is \(\mbox{D}(f(x)) = 1\) and the derivative of \(g(x) = 5\) is \(\mbox{D}(g(x)) = 0\). What are the derivatives of the elementary combinations: \(3\cdot f\), \(f + g\), \(f - g\), \(f\cdot g\) and \(\displaystyle \frac{f}{g}\)?
Solution
The first three derivatives follow “nice” patterns: \begin{align*}\mbox{D}( 3\cdot f(x) ) &= \mbox{D}( 3x ) = 3 = 3\cdot 1 = 3 \cdot \mbox{D}( f(x) )\\ \mbox{D}( f(x) + g(x) ) &= \mbox{D}( x + 5 ) = 1 = 1+0 = \mbox{D}( f(x) ) + \mbox{D}( g(x) )\\ \mbox{D}( f(x) - g(x) ) &= \mbox{D}( x - 5 ) = 1 = 1-0 = \mbox{D}( f(x) ) - \mbox{D}( g(x) )\end{align*} yet the other two derivatives fail to follow the same “nice” patterns: \(\mbox{D}( f(x) \cdot g(x) ) = \mbox{D}( 5x ) = 5\) but \(\mbox{D}( f(x) )\cdot \mbox{D}( g(x) ) = 1\cdot 0 = 0\), and \(\displaystyle \mbox{D}\left( \frac{f(x)}{g(x)}\right) = \mbox{D}\left( \frac{x}{5} \right) = \frac15\) but \(\displaystyle \frac{\mbox{D}( f(x) )}{\mbox{D}( g(x) )} =\frac10\) is undefined.
The two very simple functions in the previous example show that, in general, \(\mbox{D}( f\cdot g ) \neq \mbox{D}( f ) \cdot \mbox{D}( g )\) and \(\displaystyle \mbox{D}\left( \frac{f}{g} \right) \neq \frac{\mbox{D}( f )}{\mbox{D}( g )}\).
For \(f(x) = 6x + 8\) and \(g(x) = 2\), compute the derivatives of \(3\cdot f\), \(f+g\), \(f-g\), \(f\cdot g\) and \(\displaystyle \frac{f}{g}\).
- Answer
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\(f(x) = 6x + 8\) and \(g(x) = 2\) so \(\mbox{D}( f(x) ) = 6\) and \(\mbox{D}( g(x) ) = 0\)
\(\mbox{D}( 3\cdot f(x) ) = 3\cdot\mbox{D}( f(x) ) = 3(6) = 18\)
\(\mbox{D}( f(x) + g(x) ) = \mbox{D}( f(x) ) + \mbox{D}( g(x) ) = 6 + 0 = 6\)
\(\mbox{D}( f(x) - g(x) ) = \mbox{D}( f(x) ) - \mbox{D}( g(x) ) = 6 - 0 = 6\)
\(\mbox{D}( f(x)\cdot g(x) ) = f(x)g'(x)+g(x)f'(x) = (6x + 8)(0) + (2)(6) = 12\)
\(\mbox{D}\left( \frac{f(x)}{g(x)} \right) = \frac{g(x)f'(x) - f(x)g'(x)}{\left[ g(x) \right]^2} = \frac{(2)(6) - (6x + 8)(0)}{2^2} = \frac{12}{4} = 3\)
If: \(f\) and \(g\) are differentiable at \(x\) then:
- Constant Multiple Rule: \(\mbox{D}( k\cdot f(x) ) = k\cdot \mbox{D}( f(x) )\)
- Sum Rule: \(\mbox{D}( f(x) + g(x) ) = \mbox{D}( f(x) ) + \mbox{D}( g(x) )\)
- Difference Rule: \(\mbox{D}( f(x) - g(x) ) = \mbox{D}( f(x) ) - \mbox{D}( g(x) )\)
- Product Rule: \(\mbox{D}( f(x)\cdot g(x) ) = f(x) \cdot \mbox{D}( g(x) ) + g(x)\cdot \mbox{D}( f(x) )\)
- Quotient Rule: \(\mbox{D}\left( \frac{f(x)}{g(x)}\right) = \frac{ g(x)\cdot \mbox{D}( f(x) ) - f(x) \cdot \mbox{D}( g(x) )}{\left[g(x)\right]^2}\)
Part (e) requires that \(g(x) \neq 0\.
This theorem says that the simple patterns in the previous example for constant multiples of functions and sums and differences of functions are true for all differentiable functions. It also includes the correct patterns for derivatives of products and quotients of differentiable functions.
The proofs of parts (a), (b) and (c) of this theorem are straightforward, but parts (d) and (e) require some clever algebraic manipulations. Let’s look at some examples before tackling the proof.
Recall that \(\mbox{D}( x^2 ) = 2x\) and \(\mbox{D}( \sin(x) ) = \cos(x)\). Find \(\mbox{D}( 3\sin(x) )\) and \(\frac{d}{dx} ( 5x^2 - 7\sin(x) )\).
Solution
Computing \(\mbox{D}( 3\sin(x) )\) requires part (a) of the theorem with \(k = 3\) and \(f(x) = \sin(x)\) so \(\mbox{D}( 3\cdot \sin(x) ) = 3\cdot \mbox{D}( \sin(x) ) = 3 \cos(x)\), while \(\displaystyle \frac{d}{dx} ( 5x^2 - 7\sin(x) )\) uses part (c) of the theorem with \(f(x) = 5x^2\) and \(g(x) = 7\sin(x)\) so:
\begin{align*}\frac{d}{dx}( 5x^2 - 7\sin(x) ) &= \frac{d}{dx}( 5x^2 ) -\frac{d}{dx}( 7\sin(x) )\\
&= 5 \cdot \frac{d}{dx} ( x^2 ) - 7\cdot \frac{d}{dx} (\sin(x) )\\
&= 5( 2x ) - 7( \cos(x) )\end{align*}
which simplifies to \(10x - 7 \cos(x)\).
Find \(\mbox{D}( x^3 - 5\sin(x) )\) and \(\displaystyle \frac{d}{dx} (\sin(x) - 4x^3 )\).
- Answer
-
\(\mbox{D}( x^3- 5\sin(x) ) = \mbox{D}( x^3) - 5\cdot \mbox{D}( \sin(x) ) = 3x^2 - 5\cos(x)\)
\(\displaystyle \frac{d}{dx} \left(\sin(x) - 4x^3\right) = \frac{d}{dx} \sin(x) - 4\cdot \frac{d}{dx} x^3 = \cos(x) - 12x^2\)
The table below gives the values of functions \(f\) and \(g\), as well as their derivatives, at various points. Fill in the missing values for \(\mbox{D}( 3\cdot f(x) )\), \(\mbox{D}( 2\cdot f(x)+g(x) )\) and \(\mbox{D}( 3\cdot g(x) - f(x) )\).
| \(x\) | \(f(x)\) | \(f'(x)\) | \(g(x)\) | \(g'(x)\) | \(\mbox{D}( 3f(x) )\) | \(\mbox{D}( 2f(x) + g(x) )\) | \(\mbox{D}( 3g(x) - f(x) )\) |
|---|---|---|---|---|---|---|---|
| \(0\) | \(3\) | \(-2\) | \(-4\) | \(3\) | |||
| \(1\) | \(2\) | \(-1\) | \( 1\) | \(0\) | |||
| \(2\) | \(4\) | \( 2\) | \( 3\) | \(1\) |
- Answer
-
\(\mbox{D}( 3f(x) )\) \(\mbox{D}( 2f(x) + g(x) )\) \(\mbox{D}( 3g(x) - f(x) )\) \(-6\) \(-1\) \(11\) \(-3\) \(-2\) \(1\) \(6\) \(5\) \(1\)
Use the Main Differentiation Theorem to complete the table.
| \(x\) | \(f(x)\) | \(f'(x)\) | \(g(x)\) | \(g'(x)\) | \(\mbox{D}( f(x)\cdot g(x) )\) | \(\displaystyle \mbox{D}\left( \frac{f(x)}{g(x)} \right)\) | \(\displaystyle \mbox{D}\left( \frac{g(x)}{f(x)} \right)\) |
|---|---|---|---|---|---|---|---|
| \(0\) | \(3\) | \(-2\) | \(-4\) | \(3\) | |||
| \(1\) | \(2\) | \(-1\) | \( 1\) | \(0\) | |||
| \(2\) | \(4\) | \( 2\) | \( 3\) | \(1\) |
- Answer
-
\(\mbox{D}( f(x)\cdot g(x) )\) \(\displaystyle \mbox{D}\left( \frac{f(x)}{g(x)} \right)\) \(\displaystyle \mbox{D}\left( \frac{g(x)}{f(x)} \right)\) \(3\cdot 3+(-4)(-2) = 17\) \(\frac{-4(-2)-(3)(3)}{(-4)^2} = -\frac{1}{16}\) \(\frac{(3)(3)-(-4)(-2)}{3^2} = \frac19\) \(2\cdot 0+1(-1) = -1\) \(\frac{1(-1)-(2)(0)}{1^2} = -1\) \(\frac{2(0)-1(-1)}{2^2} = \frac14\) \(4\cdot 1 + 3\cdot 2 = 10\) \(\frac{3(2)-(4)(1)}{3^2} = \frac29\) \(\frac{4(1) - 3(2)}{4^2} = -\frac18\)
Many calculus students find it easier to remember the Product Rule in words: “the first function times the derivative of the second plus the second function times the derivative of the first.”
Determine \(\mbox{D}( x^2 \cdot \sin(x) )\) and \(\displaystyle \frac{d}{dx} \left(\frac{x^3}{\sin(x)}\right)\).
Solution
Use the Product Rule with \(f(x) = x^2\) and \(g(x) = \sin(x)\): \begin{align*} \mbox{D}( x^2\cdot \sin(x) ) &= \mbox{D}( f(x)\cdot g(x) ) = f(x)\cdot\mbox{D}( g(x) ) + g(x)\cdot \mbox{D}( f(x) )\\ &= x^2 \cdot \mbox{D}( \sin(x) ) + \sin(x)\cdot \mbox{D}( x^2)\\ &= x^2 \cdot \cos(x) + \sin(x) \cdot 2x = x^2 \cos(x) + 2x \sin(x) \end{align*} Use the Quotient Rule with \(f(x) = x^3\) and \(g(x) = \sin(x)\): \begin{align*} \frac{d}{dx} \left(\frac{x^3}{\sin(x)}\right) &= \frac{d}{dx} \left(\frac{f(x)}{g(x)}\right)\\ &= \frac{g(x)\cdot \mbox{D}( f(x) ) - f(x)\cdot \mbox{D}( g(x) )}{\left[g(x)\right]^2}\\ &= \frac{\sin(x) \cdot \mbox{D}( x^3) - x^3 \cdot \mbox{D}(\sin(x) )}{\left[ \sin(x) \right]^2}\\ &= \frac{\sin(x) \cdot 3x^2 - x^3 \cdot \cos(x)}{\sin^2(x)}\\ &= \frac{3x^2\sin(x) - x^3 \cdot \cos(x)}{\sin^2(x)} \end{align*} which could also be rewritten in terms of \(\csc(x)\) and \(\cot(x)\).
The Quotient Rule in words: “the bottom times the derivative of the top minus the top times the derivative of the bottom, all over the bottom squared.”
Find \(\mbox{D}( (x^2 + 1)(7x - 3) )\), \(\displaystyle \frac{d}{dt} \left(\frac{3t - 2}{5t + 1} \right)\) and \(\displaystyle \mbox{D}\left( \frac{\cos(x)}{x}\right)\).
- Answer
-
\(\mbox{D}((x^2 + 1)(7x - 3)) = (x^2 + 1)\mbox{D}( 7x - 3 ) + (7x - 3)\mbox{D}( x^2 + 1 ) = (x^2 + 1)(7) + (7x - 3)(2x) = 21x^2 - 6x + 7\)
or: \(\mbox{D}((x^2 + 1)(7x - 3)) = \mbox{D}( 7x^3- 3x^2 + 7x ) = 21x^2- 6x + 7\)
\(\displaystyle \frac{d}{dt} \left(\frac{3t - 2}{5t + 1} \right) = \frac{(5t + 1)\mbox{D}( 3t - 2 ) - (3t - 2)\mbox{D}( 5t + 1 )}{(5t + 1)^2} =
\frac{(5t + 1)(3) - (3t - 2)(5)}{(5t + 1)^2} = \frac{13}{(5t + 1)^2}\)\(\displaystyle \mbox{D}\left(\frac{\cos(x)}{x}\right) = \frac{x\mbox{D}( \cos(x) ) - \cos(x)\mbox{D}( x )}{x^2} = \frac{x(-\sin(x) ) - \cos(x)(1)}{x^2} = \frac{-x\cdot \sin(x) - \cos(x)}{x^2}\)
Now that we’ve seen how to use the theorem, let’s prove it.
The only general fact we have about derivatives is the definition as a limit, so our proofs here will need to recast derivatives as limits and then use some results about limits. The proofs involve applications of the definition of the derivative and results about limits.
Proof
- Using the derivative definition and the limit laws: \begin{align*}\mbox{D}( k\cdot f(x) ) &= \lim_{h\to 0} \, \frac{k \cdot f(x + h) - k \cdot f(x)}{h}\\ &= \lim_{h\to 0} \, k \cdot \frac{f(x + h) - f(x)}{h}\\ &= k \cdot \lim_{h\to 0} \, \frac{f(x + h)- f(x)}{h} = k\cdot \mbox{D}(f(x))\end{align*}
- You try it (see Practice problem that follows).
- Once again using the derivative definition and the limit laws: \begin{align*}\mbox{D}( f(x) - g(x) ) &= \lim_{h\to 0} \, \frac{\left[f(x + h)-g(x + h)\right] - \left[f(x) - g(x)\right]}{h}\\ &= \lim_{h\to 0} \, \frac{\left[f(x + h) - f(x)\right] - \left[g(x + h)-g(x)\right]}{h}\\ &= \lim_{h\to 0} \, \frac{f(x + h)- f(x)}{h} - \lim_{h\to 0}\, \frac{g(x + h)- g(x)}{h}\\ &= \mbox{D}( f(x) ) - \mbox{D}( g(x) )\end{align*}
The proofs of parts (d) and (e) of the theorem are more complicated but only involve elementary techniques, used in just the right way. Sometimes we will omit such computational proofs, but the Product and Quotient Rules are fundamental techniques you will need hundreds of times.
- By the hypothesis, \(f\) and \(g\) are differentiable, so:\[ \lim_{h\to 0} \, \frac{f(x + h)-f(x)}{h} =f'(x)\nonumber\]and:\[ \lim_{h\to 0} \, \frac{g(x + h)-g(x)}{h} =g'(x)\nonumber\]Also, both \(f\) and \(g\) are continuous (why?) so \(\displaystyle \lim_{h\to 0} \, f(x + h) = f(x)\) and \(\displaystyle \lim_{h\to 0} \, g(x + h) = g(x)\). Let \(P(x) = f(x)\cdot g(x)\). Then \(P(x+h) = f(x+h)\cdot g(x+h)\) and: \begin{align*} \mbox{D}( f(x)\cdot g(x) ) &= \mbox{D}( P(x) ) = \lim_{h\to 0} \, \frac{P(x + h)-P(x)}{h}\\ &= \lim_{h\to 0} \, \frac{f(x + h)\cdot g(x + h) - f(x)\cdot g(x)}{h}\end{align*} At this stage we need to use some cleverness to add and subtract \(f(x)\cdot g(x + h)\) from the numerator (you'll see why shortly):\[\lim_{h\to 0} \, \frac{f(x + h)\cdot g(x + h)+ \left[-f(x)\cdot g(x + h)+ f(x)\cdot g(x + h)\right] - f(x)g(x)}{h}\nonumber\]We can then split this giant fraction into two more manageable limits:\[\lim_{h\to 0} \, \frac{f(x + h)g(x + h) - f(x)g(x + h)}{h} + \lim_{h\to 0} \, \frac{f(x)g(x + h) - f(x)g(x)}{h}\nonumber\]and then factor out a common factor from each numerator:\[\lim_{h\to 0} \, g(x + h) \cdot \frac{f(x + h) - f(x)}{h} + \lim_{h\to 0} \, f(x)\cdot \frac{g(x + h) - g(x)}{h}\nonumber\]Taking limits of each piece (and using the continuity of \(g(x)\)) we get:\[\mbox{D}( f(x)\cdot g(x) ) = g(x) \cdot f'(x) + f(x) \cdot g'(x) = g \cdot \mbox{D}(f) + f\cdot \mbox{D}(g)\nonumber\]
The steps for a proof of the Quotient Rule appear in Problem 69.
Prove the Sum Rule: \(\mbox{D}( f(x) + g(x) ) = \mbox{D}( f(x) ) + \mbox{D}( g(x) )\). (Refer to the proof of part (c) for guidance.)
- Answer
-
Mimicking the proof of the Difference Rule: \begin{align*} \mbox{D}( f(x) + g(x) ) &= \lim_{h\to 0} \, \frac{\left[f(x + h)+ g(x + h)\right] - \left[f(x)+ g(x)\right]}{h}\\ &= \lim_{h\to 0} \, \frac{\left[f(x + h)- f(x)\right] + \left[g(x + h) - g(x)\right]}{h}\\ &= \lim_{h\to 0} \, \frac{f(x + h)- f(x)}{h} + \lim_{h\to 0} \, \frac{g(x + h)- g(x)}{h}\\ &= \mbox{D}( f(x) ) + \mbox{D}( g(x) )\end{align*}
Using the Differentiation Rules
You definitely need to memorize the differentiation rules, but it is vitally important that you also know how to use them. Sometimes it is clear that the function we want to differentiate is a sum or product of two obvious functions, but we commonly need to differentiate functions that involve several operations and functions. Memorizing the differentiation rules is only the first step in learning to use them.
Calculate \(\mbox{D}( x^5 + x\cdot \sin(x) )\).
Solution
This function is more difficult because it involves both an addition and a multiplication. Which rule(s) should we use — or, more importantly, which rule should we use first?
First apply the Sum Rule to trade one derivative for two easier ones:
\begin{align*}
\mbox{D}( x^5 + x\cdot\sin(x) ) &= \mbox{D}( x^5) + \mbox{D}( x\cdot \sin(x) )\\ &= 5x^4 + \left[x\cdot \mbox{D}( \sin(x) ) + \sin(x)\cdot \mbox{D}( x )\right]\\ &= 5x^4 + x\cdot\cos(x) + \sin(x)
\end{align*}
This last expression involves no more derivatives, so we are done.
If instead of computing the derivative you were evaluating the function \(x^5+ x \sin(x)\) for some particular value of \(x\), you would:
- raise \(x\) to the 5th power
- calculate \(\sin(x)\)
- multiply \(\sin(x)\) by \(x\) and, finally,
- add (sum) the values of \(x^5\) and \(x \sin(x)\)
Notice that the final step of your evaluation of \(f\) indicates the first rule to use to calculate the derivative of \(f\).
Which differentiation rule should you apply first for each of the following?
- \(x\cdot\cos(x) - x^3\cdot\sin(x)\)
- \((2x - 3)\cos(x)\)
- \(2\cos(x) - 7x^2\)
- \(\displaystyle \frac{\cos(x) + 3x}{\sqrt{x}}\)
- Answer
-
- difference rule
- product rule
- difference rule
- quotient rule
Calculate \(\displaystyle \mbox{D}\left( \frac{x^2 - 5}{\sin(x)}\right)\) and \(\displaystyle \frac{d}{dt} \left(\frac{t^2 - 5}{t\cdot \sin(t)}\right)\).
- Answer
-
\(\displaystyle \mbox{D}\left({x^2- 5}{\sin(x)}\right) = \frac{\sin(x)\mbox{D}(x^2- 5) - (x^2- 5)\mbox{D}(\sin(x))}{(\sin(x))^2} = \frac{\sin(x)(2x) - (x^2- 5)\cos(x)}{\sin^2(x)}\)
\(\displaystyle\frac{d}{dt}\left({t^2- 5}{t\cdot\sin(t)}\right) = \frac{t\cdot\sin(t)\mbox{D}(t^2-5) - (t^2-5)\mbox{D}(t\cdot\sin(t))}{(t\cdot\sin(t))^2} = \frac{t\cdot\sin(t)(2t) - (t^2- 5)\left[ t\cos(t) + \sin(t)\right]}{t^2\cdot\sin^2(t)}\)
A mass attached to a spring oscillates up and down but the motion becomes “damped” due to friction and air resistance. The height of the mass after \(t\) seconds is given by \(\displaystyle h(t) = 5 + \frac{\sin(t)}{1 + t}\) (in feet).
Find the height and velocity of the mass after \(2\) seconds.
Solution
The height is \(\displaystyle h(2) = 5 + \frac{\sin(2)}{1 + 2} \approx 5 + \frac{0.909}{3} = 5.303\) feet above the ground. The velocity is \(h '(2)\), so we must first compute \(h'(t)\) and then evaluate the derivative at time \(t=2\):\[h'(t) = \frac{(1 + t) \cdot \cos(t) - \sin(t)\cdot 1}{(1 + t)^2}\)\nonumber\]so \(\displaystyle h'(2) = \frac{3\cos(2) - \sin(2)}{9} \approx \frac{-2.158}{9} \approx -0.24\) feet per second.
What are the height and velocity of the weight in Example \(\PageIndex{8}\) after \(5\) seconds? What are the height and velocity of the weight be after a “long time” has passed?
- Answer
-
\(h(5) = 5 + \frac{\sin(5)}{1+5} \approx 4.84\) ft.; \(v(5) = h'(5) = \frac{(1+5)\cos(5) - \sin(5)}{(1+5)^2} \approx 0.074\) ft/sec
“long time”: \(h(t) = 5 + \frac{\sin(t)}{1 + t} \approx 5\) feet when \(t\) is very large
\(\displaystyle h'(t) = \frac{(1+ t)\cos(t) - \sin(t)}{(1+ t)^2} = \frac{\cos(t)}{1+t} - \frac{\sin(t)}{(1+t)^2} \approx 0\) ft/sec when \(t\) is very large
Calculate \(\mbox{D}( x\cdot\sin(x)\cdot \cos(x) )\).
Solution
Clearly we need to use the Product Rule, because the only operation in this function is multiplication. But the Product Rule deals with a product of two functions and here we have the product of three: \(x\) and \(\sin(x)\) and \(\cos(x)\). If, however, we think of our two functions as \(f(x) = x\cdot \sin(x)\) and \(g(x) = \cos(x)\), then we do have the product of two functions and:
\begin{align*}
\mbox{D}( x \cdot \sin(x)\cdot\cos(x) ) &= \mbox{D}( f(x)\cdot g(x) )\\ &= f(x) \cdot \mbox{D}( g(x) ) + g(x)\cdot \mbox{D}( f(x) )\\
&= x \sin(x) \cdot \mbox{D}( \cos(x) ) + \cos(x)\cdot \mbox{D}( x \sin(x) )
\end{align*}
We are not done, but we have traded one hard derivative for two easier ones. We know that \(\mbox{D}( \cos(x) ) = -\sin(x)\) and we can use the Product Rule (again) to calculate \(\mbox{D}( x \sin(x) )\). Then the
last line of our calculation above becomes:\[x \sin(x)\cdot\left[ -\sin(x) \right] + \cos(x)\cdot\left[ x \mbox{D}( \sin(x) ) + \sin(x) \mbox{D}( x )\right]\nonumber\]
and then:\[-x \sin^2(x) + \cos(x) \left[ x \cos(x) + \sin(x) (1)\right]\nonumber\]which simplifies to \(-x \sin^2(x) + x \cos^2(x) + \cos(x) \sin(x)\).
Evaluating a Derivative at a Point
The derivative of a function \(f(x)\) is a new function \(f'(x)\) that tells us the slope of the line tangent to the graph of \(f\) at each point \(x\). To find the slope of the tangent line at a particular point \(( c, f(c) )\) on the graph of \(f\), we should first calculate the derivative \(f'(x)\) and then evaluate the function \(f'(x)\) at the point \(x = c\) to get the number \(f'(c)\). If you mistakenly evaluate \(f\) first, you get a number \(f(c)\), and the derivative of a constant is always equal to \(0\).
Determine the slope of the line tangent to the graph of \(f(x) = 3x + \sin(x)\) at \((0, f(0) )\) and \((1, f(1 ))\).
Solution
\(f'(x) = \mbox{D}( 3x + \sin(x) ) = \mbox{D}(3x) + \mbox{D}( \sin(x) ) = 3 + \cos(x)\). When \(x = 0\), the graph of \(y = 3x + \sin(x)\) goes through the point \(( 0, 3(0)+\sin(0) ) = (0,0)\) with slope \(f'(0) = 3 + \cos(0) = 4\). When \(x = 1\), the graph goes through the point \(( 1, 3(1)+\sin(1) ) \approx (1, 3.84)\) with slope \(f'(1) = 3 + \cos(1) \approx 3.54\).
Where do \(f(x) = x^2- 10x + 3\) and \(g(x) = x^3 - 12x\) have horizontal tangent lines?
- Answer
-
\(f'(x) = 2x - 10\) so \(f'(x) = 0 \Rightarrow 2x - 10 = 0 \Rightarrow x = 5\)
\(g'(x) = 3x^2 - 12\) so \(g'(x) = 0 \Rightarrow 3x^2 - 12 = 0 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2\)
Important Information and Results
This section, like the last one, contains a great deal of important information that we will continue to use throughout the rest of the course, so we collect here some of those important results.
Differentiability and Continuity: If a function is differentiable then it must be continuous. If a function is not continuous then it cannot be differentiable. A function may be continuous at a point and not differentiable there.
Graphically: Continuous means “connected”; differentiable means “continuous, smooth and not vertical.”
Differentiation Patterns:
- \(\left[k\cdot f(x)\right]' = k\cdot f'(x)\)
- \(\left[f(x)+g(x)\right]' = f'(x) +g'(x)\)
- \(\left[f(x)-g(x)\right]' = f'(x) -g'(x)\)
- \(\left[f(x)\cdot g(x)\right]' = f(x)\cdot g'(x) + g(x)\cdot f'(x)\)
- \(\displaystyle \left[\frac{f(x)}{g(x)}\right]' = \frac{g(x)\cdot f'(x) - f(x)\cdot g'(x)}{\left[g(x)\right]^2}\)
- The final step used to evaluate a function \(f\) indicates the first rule used to differentiate \(f\).
Evaluating a derivative at a point: First differentiate and then evaluate.
Problems
- Use the graph of \(y = f(x)\) below to determine:
- at which integers \(f\) is continuous.
- at which integers \(f\) is differentiable.
- Use the graph of \(y = g(x)\) below to determine:
- at which integers \(g\) is continuous.
- at which integers \(g\) is differentiable.
- Use the values given in the table to determine the values of \(f\cdot g\), \(\mbox{D}(f\cdot g )\), \(\frac{f}{g}\) and \(\mbox{D}\left(\frac{f}{g}\right)\).
\(x\) \(f(x)\) \(f'(x)\) \(g(x)\) \(g'(x)\) \(f(x)\cdot g(x)\) \(\mbox{D}( f(x)\cdot g(x) )\) \(\frac{f(x)}{g(x)}\) \(\mbox{D}\left( \frac{f(x)}{g(x)} \right)\) \(0\) \(2\) \(3\) \(1\) \(5\) \(1\) \(-3\) \(2\) \(5\) \(-2\) \(2\) \(0\) \(-3\) \(2\) \(4\) \(3\) \(1\) \(-1\) \(0\) \(3\) - Use the values given in the table to determine the values of \(f\cdot g\), \(\mbox{D}(f\cdot g )\), \(\frac{f}{g}\) and \(\mbox{D}\left(\frac{f}{g}\right)\).
\(x\) \(f(x)\) \(f'(x)\) \(g(x)\) \(g'(x)\) \(f(x)\cdot g(x)\) \(\mbox{D}( f(x)\cdot g(x) )\) \(\frac{f(x)}{g(x)}\) \(\mbox{D}\left( \frac{f(x)}{g(x)} \right)\) \(0\) \(4\) \(2\) \(3\) \(-3\) \(1\) \(0\) \(3\) \(2\) \(1\) \(2\) \(-2\) \(5\) \(0\) \(-1\) \(3\) \(-1\) \(-2\) \(-3\) \(4\) - Use the information in the figure below left to plot the values of the functions \(f + g\), \(f\cdot g\) and \(\frac{f}{g}\) and their derivatives at \(x = 1\), \(2\) and \(3\).
![On the left is a blue graph of y = f(x) on the grid [0,3]X[0,3]. It rises roughly linearly through (0,3) to a peak near (0.8,3.5) and siwtched from concave down to concave up at (1,3), reaching a low point at (2,1) and rising roughly linearly through (4,2). On the right is a red graoh of y=g(x) on a [0,3]X[0,3] grid that rises linearly through (0,1) before bending downward slightly but still increasing to (0.7,0.5), where it becomes concave up and then linear through (1,2), becoming concave down near ((1.2,2.7), reaching a peak near (1.4,2.8), becoming concave up at x = 1.5, reaching a low point at x = 1.6 and rising through (2,3), where it becomes concave down, reaches a peak at (2.4,3.3) and then descends through (3,2.4).](https://math.libretexts.org/@api/deki/files/140188/fig202_8.png?revision=1&size=bestfit&width=585&height=244)
- Use the information in the figure above right to plot the values of the functions \(2f\), \(f - g\) and \(\frac{g}{f}\) and their derivatives at \(x = 1\), \(2\) and \(3\).
- Calculate \(\mbox{D}( (x - 5)(3x + 7) )\) by:
- using the Product Rule.
- expanding and then differentiating.
- Calculate \(\displaystyle \mbox{D}\left( \frac{x^3-3x+2}{\sqrt{x}} \right)\) by:
- using the Product Rule.
- expanding and then differentiating.
In Problems 9–26, compute each derivative.
- \(\displaystyle \frac{d}{dx} \left(19x^3 -7\right)\)
- \(\displaystyle \frac{d}{dt} \left(5\cos(t) +\frac{\pi}{2}\right)\)
- \(\mbox{D}( \sin(x) + \cos(x) )\)
- \(\mbox{D}( 7\sin(x) - 3\cos(x) )\)
- \(\mbox{D}( x^2\cdot \cos(x) )\)
- \(\mbox{D}( \sqrt{x}\cdot \sin(x) )\)
- \(\mbox{D}( \sin^2(x) )\)
- \(\mbox{D}( \cos^2(x) )\)
- \(\displaystyle \frac{d}{dx} \left(\frac{\cos(x)}{x^2} \right)\)
- \(\displaystyle \frac{d}{dt} \left(\frac{\sin(t)}{t^3} \right)\)
- \(\displaystyle \frac{d}{dx} \left(\frac{1}{1+x^2} \right)\)
- \(\displaystyle \frac{d}{dt} \left(\frac{t}{1+t^3} \right)\)
- \(\displaystyle \frac{d}{d\theta} \left(\frac{1}{\cos(\theta)} \right)\)
- \(\displaystyle \frac{d}{d\theta} \left(\frac{1}{\sin(\theta)} \right)\)
- \(\displaystyle \frac{d}{d\theta} \left(\frac{\sin(\theta)}{\cos(\theta)} \right)\)
- \(\displaystyle \frac{d}{d\theta} \left(\frac{\cos(\theta)}{\sin(\theta)} \right)\)
- \(\displaystyle \mbox{D}\left(8x^5-3x^4+2x^3+7x^2-12x+147\right)\)
-
- \(\mbox{D}( \sin(x) )\)
- \(\displaystyle \mbox{D} \left(\sin(x) + 7 \right)\)
- \(\mbox{D}(\sin(x) - 8000 )\)
- \(\mbox{D}( \sin(x) + k )\)
- Find values for the constants \(a\), \(b\) and \(c\) so that the parabola \(f(x) = ax^2 + bx + c\) has \(f(0) = 0\), \(f'(0) = 0\) and \(f'(10) = 30\).
- If \(f\) is a differentiable function, how are the:
- graphs of \(y = f(x)\) and \(y = f(x) + k\) related?
- derivatives of \(f(x)\) and \(f(x) + k\) related?
- If \(f\) and \(g\) are differentiable functions that always differ by a constant (\)f(x) - g(x) = k\) for all \(x\)) then what can you conclude about their graphs? Their derivatives?
- If \(f\) and \(g\) are differentiable functions whose sum is a constant (\)f(x) + g(x) = k\) for all \(x\)) then what can you conclude about their graphs? Their derivatives?
- If the product of \(f\) and \(g\) is a constant (that is, \(f(x)\cdot g(x) = k\) for all \(x\)) then how are \(\displaystyle \frac{\mbox{D}( f(x) )}{f(x)}\) and \(\displaystyle \frac{\mbox{D}( g(x) )}{g(x)}\) related?
- If the quotient of \(f\) and \(g\) is a constant (\(\displaystyle \frac{f(x)}{g(x)} = k\) for all \(x\)) then how are \(g\cdot f'\) and \(f\cdot g'\) related?
In Problems 33–40:
- calculate \(f '(1)\)
- determine where \(f'(x) = 0\).
- \(f(x) = x^2- 5x + 13 \)
- \(f(x) = 5x^2- 40x + 73\)
- \(f(x) = 3x - 2\cos(x)\)
- \(f(x) = \left| x + 2 \right|\)
- \(f(x) = x^3 + 9x^2 + 6 \)
- \(f(x) = x^3 + 3x^2 + 3x - 1\)
- \(f(x) = x^3 + 2x^2 + 2x - 1\)
- \(\displaystyle f(x) = \frac{7x}{x^2 + 4}\)
- \(f(x) = x\cdot \sin(x)\) and \(0 \leq x \leq 5\). (You may need to use the Bisection Algorithm or the “trace” option on a calculator to approximate where \(f'(x) = 0\).)
- \(f(x) = Ax^2 + B x + C\), where \(B\) and \(C\) are constants and \(A \neq 0\) is constant.
- \(f(x) = x^3 + A x^2 + B x + C\) with constants \(A\), \(B\) and \(C\). Can you find conditions on the constants \(A\), \(B\) and \(C\) that will guarantee that the graph of \(y = f(x)\) has two distinct “turning points”? (Here a “turning point” means a place where the curve changes from increasing to decreasing or from decreasing to increasing, like the vertex of a parabola.)
In Problems 44–51, where are the functions differentiable?
- \(f(x) = \left|x\right| \cos(x)\)
- \(f(x) = \tan(x)\)
- \(\displaystyle f(x) = \frac{x - 5}{x + 3}\)
- \(\displaystyle f(x) = \frac{x^2 + x}{x^2- 3x}\)
- \(f(x) = \left| x^2 - 4 \right|\)
- \(f(x) = \left| x^3 - 1 \right|\)
- \(\displaystyle f(x) = \left\{ \begin{array}{rl} { 0 } & { \text{if } x < 0 } \\ { \sin(x) } & { \text{if } x \geq 0 } \end{array}\right.\)
- \(\displaystyle f(x) = \left\{ \begin{array}{rl} { x } & { \text{if } x < 0 } \\ { \sin(x) } & { \text{if } x \geq 0 } \end{array}\right.\)
- For what value(s) of \(A\) is \(f(x) = \left\{ \begin{array}{rl} { Ax-4 } & { \text{if } x < 2 } \\ { x^2+x } & { \text{if } x \geq 2 } \end{array}\right.\) differentiable at \(x = 2\)?
- For what values of \(A\) and \(B\) is \(f(x) = \left\{ \begin{array}{rl} { Ax+B } & { \text{if } x < 1 } \\ { x^2+x } & { \text{if } x \geq 1 }\end{array}\right.\) differentiable at \(x = 1\)?
- An arrow shot straight up from ground level (get out of the way!) with an initial velocity of \(128\) feet per second will be at height \(h(x) = -16x^2 + 128x\) feet after \(x\) seconds (see figure below).
- Determine the velocity of the arrow when \(x = 0\), \(1\) and \(2\) seconds.
- What is the velocity of the arrow, \(v(x)\), at any time \(x\)?
- At what time \(x\) will the velocity of the arrow be \(0\)?
- What is the greatest height the arrow reaches?
- How long will the arrow be aloft?
- Use the answer for the velocity in part (b) to determine the acceleration, \(a(x) = v'(x)\), at any time \(x\).
- If an arrow is shot straight up from ground level on the moon with an initial velocity of \(128\) feet per second, its height will be \(h(x) = -2.65x^2 + 128x\) feet after \(x\) seconds. Redo parts (a)–(e) of problem 40 using this new formula for \(h(x)\).
- In general, if an arrow is shot straight upward with an initial velocity of \(128\) feet per second from ground level on a planet with a constant gravitational acceleration of \(g\) feet per second\)^2\) then its height will be \(h(x) = -\frac{g}{2}x^2 + 128x\) feet after \(x\) seconds. Answer the questions in problem 40 for arrows shot on Mars and Jupiter.
object \(g\) (ft/sec\(^2\)) \(g\) (cm/sec\(^2\)) Mercury 11.8 358 Venus 20.1 887 Earth 32.2 981 moon 5.3 162 Mars 12.3 374 Jupiter 85.3 2601 Saturn 36.6 1117 Uranus 34.4 1049 Neptune 43.5 1325 Source: CRC Handbook of Chemistry and Physics - If an object on Earth is propelled upward from ground level with an initial velocity of \(v_0\) feet per second, then its height after \(x\) seconds will be \(h(x) = -16x^2 +v_0 x\).
- Find the object's velocity after \(x\) seconds.
- Find the greatest height the object will reach.
- How long will the object remain aloft?
- In order for a 6-foot-tall basketball player to dunk the ball, the player must achieve a vertical jump of about 3 feet. Use the information in the previous problems to answer the following questions.
- What is the smallest initial vertical velocity the player can have and still dunk the ball?
- With the initial velocity achieved in part (a), how high would the player jump on the moon?
- The best high jumpers in the world manage to lift their centers of mass approximately 3.75 feet.
- What is the initial vertical velocity these high jumpers attain?
- How long are these high jumpers in the air?
- How high would they lift their centers of mass on the moon?
-
- Find an equation for the line \(L\) that is tangent to the curve \(\displaystyle y = \frac{1}{x}\) at the point \((1,1)\).
- Determine where \(L\) intersects the \(x\)-axis and the \(y\)-axis.
- Determine the area of the region in the first quadrant bounded by \(L\), the \(x\)-axis and the \(y\)-axis (see figure below).
-
- Find an equation for the line \(L\) that is tangent to the curve \(\displaystyle y = \frac{1}{x}\) at the point \((2, \frac12)\).
- Graph \(y = \frac{1}{x}\) and \(L\) and determine where \(L\) intersects the \(x\)-axis and the \(y\)-axis.
- Determine the area of the region in the first quadrant bounded by \(L\), the \(x\)-axis and the \(y\)-axis.
-
- Find an equation for the line \(L\) that is tangent to the curve \(\displaystyle y = \frac{1}{x}\) at the point \((p, \frac{1}{p})\) (assuming \(p \neq 0\)).
- Determine where \(L\) intersects the \(x\)-axis and the \(y\)-axis.
- Determine the area of the region in the first quadrant bounded by \(L\), the \(x\)-axis and the \(y\)-axis.
- How does the area of the triangle in part (c) depend on the initial point \((p, \frac{1}{p})\)?
- Find values for the coefficients \(a\), \(b\) and \(c\) so that the parabola \(f(x) = ax^2 + bx + c\) goes through the point \((1,4)\) and is tangent to the line \(y = 9x - 13\) at the point \((3, 14)\).
- Find values for the coefficients \(a\), \(b\) and \(c\) so that the parabola \(f(x) = ax^2 + bx + c\) goes through the point \((0,1)\) and is also tangent to the line \(y = 3x - 2\) at the point \((2,4)\).
-
- Find a function \(f\) so that \(\mbox{D}( f(x) ) = 3x^2\).
- Find another function \(g\) with \(\mbox{D}( g(x) ) = 3x^2\).
- Can you find more functions whose derivatives are \(3x^2\)?
-
- Find a function \(f\) so that \(f'( x ) = 6x + \cos(x)\).
- Find another function \(g\) with \(g'( x ) = f'(x)\).
- The graph of \(y = f'(x)\) appears below.
- Assume \(f(0) = 0\) and sketch a graph of \(y = f(x)\).
- Assume \(f(0) = 1\) and graph \(y = f(x)\).
- The graph of \(y = g'(x)\) appears below. Assume that \(g\) is continuous.
- Assume \(g(0) = 0\) and sketch a graph of \(y = g(x)\).
- Assume \(g(0) = 1\) and graph \(y = g(x)\).
- Assume that \(f\) and \(g\) are differentiable functions and that \(g(x) \neq 0\). State why each step in the following proof of the Quotient Rule is valid.
Proof of the Quotient Rule
\begin{align*} \mbox{D}\left(\frac{f(x)}{g(x)}\right) & = \lim_{h\to 0} \, \frac{1}{h} \left[\frac{f(x+h)}{g(x+h)} - \frac{f(x)}{g(x)}\right] = \lim_{h\to 0} \, \frac{1}{h}\left[\frac{f(x+h)g(x) - g(x+h)f(x)}{g(x+h)g(x)}\right]\\ & = \lim_{h\to 0} \, \frac{1}{g(x+h)g(x)}\left[\frac{f(x+h)g(x) +(-f(x)g(x) + f(x)g(x))- g(x+h)f(x)}{h}\right]\\ & = \lim_{h\to 0} \, \frac{1}{g(x+h)g(x)}\left[g(x)\frac{f(x+h)-f(x)}{h} + f(x)\frac{g(x)- g(x+h)}{h}\right]\\ & = \frac{1}{\left[g(x)\right]^2}\left[g(x)\cdot f'(x) - f(x)\cdot g'(x)\right]\\ & = \frac{g(x)\cdot f'(x) - f(x)\cdot g'(x)}{\left[g(x)\right]^2} \end{align*}