7.2: An Alternate Definition of Derivative
- Page ID
- 178856
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)To find the slope of a tangent line to the curve \(y=f(x)\) at \((a,f(a))\) we use the derivative limit \(\textcolor{red}{\displaystyle{f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}}}\)
No matter what type of function \(f(x)\) seems to be, this limit always seems to end up with factoring the top and reducing away \((x-a)\).
Unfortunately, if your factoring skills aren't great (or \(a\) isn't very nice), this may be quite challenging for certain functions or values of \(a\).
Consider \(f(x)=x^2\) and find \(f'(\sqrt{3})\)
Solution
\(\displaystyle{f'(\sqrt{3})=\lim_{x\to \sqrt{3}} \frac{f(x)-f(\sqrt{3})}{x-\sqrt{3}}=\lim_{x\to \sqrt{3}} \frac{x^2-3}{x-\sqrt{3}}}\)
(factoring the top isn't obvious, but its possible!)
\(\displaystyle{=\lim_{x\to \sqrt{3}} \frac{(x-\sqrt{3})(x+\sqrt{3})}{x-\sqrt{3}}=\lim_{x\to \sqrt{3}} (x+\sqrt{3}) =2\sqrt{3}}\)
Clearly if you cannot factor the numerator, you won't be able to reach this answer.
Instead, we can do a little substitution inside of the derivative formula to avoid this issue:
Define \(h=x-a\), which can be re-written as \(x=a+h\).
So \(f(x)\) becomes \(f(a+h)\) and \(\displaystyle{\lim_{x\to a} h=0}\) so \(x\to a\) can be replaced by \(h\to 0\)
Putting this all together we get \(\textcolor{red}{\displaystyle{ f'(a)=\lim_{h\to 0} \frac{f(a+h)-f(a)}{h}}}\)
If we try to find \(f'(\sqrt{3})\) as we did above with this new definition:
Reconsider \(f(x)=x^2\) and find \(f'(\sqrt{3})\)
Solution
\(\displaystyle{f'(\sqrt{3})=\lim_{h\to 0} \frac{f(\sqrt{3}+h)-f(\sqrt{3})}{h}=\lim_{h\to 0} \frac{(\sqrt{3}+h)^2-3}{h}=\lim_{h\to 0} \frac{(3+2h\sqrt{3}+h^2)-3}{h}}\)
\(\displaystyle{=\lim_{h\to 0} \frac{2h\sqrt{3}+h^2}{h}=\lim_{h\to 0} (2\sqrt{3}+h) =2\sqrt{3}}\)
This alternative limit definition of the derivative will always yield the same answer as the original definition.
Pick whichever one is easiest for you to use on a given question.
Lets just try a few example problems using this new definition (the old one works fine if you prefer it, but we will get some practice using this new one)
Find the line tangent to \(g(x)=\displaystyle{\frac{x+1}{x-1}}\) at the point \(\displaystyle{ \left(\frac{1}{2},-3\right)}\)
Solution
\(\displaystyle{ g'\left( \frac{1}{2}\right) = \lim_{h\to 0} \frac{g\left(\frac{1}{2}+h\right)-g\left(\frac{1}{2}\right)}{h} = \lim_{h\to 0} \frac{ \displaystyle{ \frac{ \frac{1}{2}+h+1}{\frac{1}{2}+h-1}-(-3)}}{h}= \lim_{h\to 0} \frac{ \displaystyle{ \frac{ h+\frac{3}{2}}{h-\frac{1}{2}}+3}}{h} \cdot \textcolor{brown}{\frac{h-\frac{1}{2}}{h-\frac{1}{2}}}} \)
\(\displaystyle{ = \lim_{h\to 0} \frac{ \left(h+\frac{3}{2}\right)+3\left(h-\frac{1}{2}\right)}{h\left(h-\frac{1}{2}\right)} = \lim_{h\to 0} \frac{ 4h}{h\left(h-\frac{1}{2}\right)} = \lim_{h\to 0} \frac{ 4}{h-\frac{1}{2}} = \frac{ 4}{0-\frac{1}{2}}} =-8\)
We plug this into the tangent line formula \(\displaystyle{y-g\left(\frac{1}{2}\right)=g'\left(\frac{1}{2}\right)\left(x-\frac{1}{2}\right)} \)
Which gives our final answer, \(\displaystyle{ y+3=-8\left(x-\frac{1}{2}\right)}\)
A quick glance at the graph of makes this look believable.
The same tricks works when dealing with radicals for both versions of the derivative limit:
Find the line tangent to \(f(x)= \sqrt{x^2-x-2}\) at the point where \(x=3\)
Solution
First we see that \(f(3)=\sqrt{3^2-3-2}=\sqrt{4}=2\), now we find \(f'(3)\):
\(\displaystyle{ f'\left( 3\right) = \lim_{h\to 0} \frac{f\left(3+h\right)-f\left(3\right)}{h} = \lim_{h\to 0} \frac{ \sqrt{(3+h)^2-(3+h)-2}-2}{h}}\)
We rationalize our numerator to help us reduce this:
\(\displaystyle{ = \lim_{h\to 0} \frac{\sqrt{h^2+5h+4}-2}{h} \cdot \textcolor{purple}{\frac{\sqrt{h^2+5h+4}+2}{\sqrt{h^2+5h+4}+2}} = \lim_{h\to 0} \frac{ \left(\sqrt{h^2+5h+4}\right)^2-2^2}{h \left(\sqrt{h^2+5h+4}+2\right)} = \lim_{h\to 0} \frac{ h^2+5h+4 -4}{h \left(\sqrt{h^2+5h+4}+2\right)}} \)
\(\displaystyle{ = \lim_{h\to 0} \frac{ h^2+5h}{h \left(\sqrt{h^2+5h+4}+2\right)} = \lim_{h\to 0} \frac{ h+5}{ \sqrt{h^2+5h+4}+2} = \frac{ 0+5}{\sqrt{0^2+5\cdot 0+4}+2}} =\frac{5}{4}\)
We plug this into the tangent line formula \(\displaystyle{y-f\left(3\right)=f'\left(3\right)\left(x-3\right)}\)
Which gives our final answer, \(\displaystyle{ y-2=\frac{5}{4}\left(x-3\right)}\)
Again here is some graphical evidence that we are correct:
This definition helps to avoid factoring, but be warned that evaluating \(f(a+h)\) usually takes an equivalent amount of work (but many students think its easier work):
Find \(s'(1-\sqrt{2})\) for \(s(t)=\displaystyle{ \frac{1}{(1-t)^3}}\)
Solution
First we find \(\displaystyle{s(1-\sqrt{2})=\frac{1}{(1-(1-\sqrt{2}))^3}=\frac{1}{(\sqrt{2})^3}=\frac{1}{2\sqrt{2}}}\), now we find \(s'(1-\sqrt{2})\):
\(\displaystyle{ s'\left( 1-\sqrt{2}\right) = \lim_{h\to 0} \frac{s\left(1-\sqrt{2}+h\right)-s\left(1-\sqrt{2}\right)}{h} = \lim_{h\to 0} \frac{ \frac{1}{(1-(1-\sqrt{2}+h))^3}-\frac{1}{2\sqrt{2}}}{h}}\)
\(\displaystyle{ = \lim_{h\to 0} \frac{ \frac{1}{(\sqrt{2}-h)^3}-\frac{1}{2\sqrt{2}}}{h} = \lim_{h\to 0} \frac{ \frac{1}{(\sqrt{2})^3-3(\sqrt{2})^2\cdot h+3\sqrt{2}\cdot h^2-h^3}-\frac{1}{2\sqrt{2}}}{h}}\)
We multiply the top and bottom by the common denominator of the smaller fractions:
\(\displaystyle{ = \lim_{h\to 0} \frac{ \frac{1}{2\sqrt{2}-6h+3h^2\sqrt{2}-h^3}-\frac{1}{2\sqrt{2}}}{h} \cdot \textcolor{purple}{\frac{ 2\sqrt{2}(2\sqrt{2}-6h+3h^2\sqrt{2}-h^3) }{2\sqrt{2}(2\sqrt{2}-6h+3h^2\sqrt{2}-h^3)}}}\)
\(\displaystyle{ = \lim_{h\to 0} \frac{ 2\sqrt{2}-(2\sqrt{2}-6h+3h^2\sqrt{2}-h^3)}{2h\sqrt{2}(2\sqrt{2}-6h+3h^2\sqrt{2}-h^3)} = \lim_{h\to 0} \frac{ 6h-3h^2\sqrt{2}+h^3}{2h\sqrt{2}(2\sqrt{2}-6h+3h^2\sqrt{2}-h^3)} }\)
\(\displaystyle{ = \lim_{h\to 0} \frac{ 6-3h\sqrt{2}+h^2}{2\sqrt{2}(2\sqrt{2}-6h+3h^2\sqrt{2}-h^3)}= \frac{ 6-3\cdot 0\cdot\sqrt{2}+0^2}{2\sqrt{2}(2\sqrt{2}-6\cdot 0+3\cdot 0^2\cdot \sqrt{2}-0^3)} = \frac{3}{4} }\)
This looks ugly, but consider that if you try the \(\displaystyle{\lim_{t\to 1-\sqrt{2}}}\) version of derivative for this question, at some point you would have had to factor the term \((t-(1-\sqrt{2}))\) out of the cubic polynomial \(-t^3+3t^2-3t+1-2\sqrt{2}\)
It is always good to have an alternative approach when factoring is hard!