2.5: Applications of the Chain Rule

The Chain Rule can help us determine the derivatives of logarithmic functions like \(f(x) = \ln(x)\) and general exponential functions like \(f(x) = a^x\). We will also use it to answer some applied questions and to find slopes of graphs given by parametric equations.

Derivatives of Logarithms

You know from precalculus that the natural logarithm \(\ln(x)\) is defined as the inverse of the exponential function \(e^x\): \(\displaystyle e^{\ln(x)} = x\) for \(x > 0\). We can use this identity along with the Chain Rule to determine the derivative of the natural logarithm.

Theorem

\(\displaystyle \mbox{D}( \ln(x) ) = \frac{1}{x}\) and \(\displaystyle \mbox{D}\left( \ln( g(x) ) \right) = \frac{g'(x)}{g(x)}\)

Proof

We know that \(\mbox{D}( e^u) = e^u\), so using the Chain Rule we have \(\displaystyle \mbox{D}\left( e^{f(x)} \right) = e^{f(x)} \cdot f'(x)\). Differentiating each side of the identity \(\displaystyle e^{\ln(x)} = x\), we get:
\begin{align*}
\mbox{D}\left( e^{\ln(x)} \right) = \mbox{D}( x ) &\Rightarrow e^{\ln(x)} \cdot \mbox{D}( \ln(x) ) = 1\\
&\Rightarrow x \cdot \mbox{D}( \ln(x) ) = 1 \Rightarrow \mbox{D}( \ln(x) ) = \frac{1}{x}\end{align*}
The function \(\ln( g(x) )\) is the composition of \(f(x) = \ln(x)\) with \(g(x)\) so the Chain Rule says:\[\mbox{D}\left( \ln( g(x) \right) = \mbox{D}\left( f( g(x) ) \right) = f'( g(x) )\cdot g'(x) = \frac{1}{g(x)} \cdot g'(x) = \frac{g'(x)}{g(x)}\nonumber\]Graph \(f(x) = \ln(x)\) along with \(\displaystyle f'(x) = \frac{1}{x}\) and compare the behavior of the function at various points with the values of its derivative at those points. Does \(\displaystyle y = \frac{1}{x}\) possess the properties you would expect to see from the derivative of \(f(x) = \ln(x)\)?

You can remember the differentiation pattern for the the natural logarithm in words as: “one over the inside times the the derivative of the inside.”

Example \(\PageIndex{1}\)

Find \(\mbox{D}( \ln( \sin(x) ) )\) and \(\mbox{D}( \ln( x^2 + 3 ) )\).

Solution

Using the pattern \(\displaystyle \mbox{D}( \ln( g(x) ) = \frac{g'(x)}{g(x)}\) with \(g(x) = \sin(x)\):\[\mbox{D}( \ln( \sin(x) ) ) = \frac{g'(x)}{g(x)} =  \frac{\mbox{D}( \sin(x) )}{\sin(x)} = \frac{\cos(x)}{\sin(x)} = \cot(x)\nonumber\]With \(g(x) = x^2 + 3\), \(\displaystyle \mbox{D}( \ln( x^2 + 3 ) ) = \frac{g'(x)}{g(x)} =  \frac{2x}{x^2 + 3}\).

We can use the Change of Base Formula from precalculus to rewrite any logarithm as a natural logarithm, and then we can differentiate the resulting natural logarithm.

Change of Base Formula for Logarithms

\(\displaystyle \log_a(x) = \frac{\log_b(x)}{\log_b(a)}\) for all positive \(a\), \(b\) and \(x\).

Note

Your calculator likely has two logarithm buttons: ln for the natural logarithm (base \(e\)) and log for the common logarithm (base \(10\)). Be careful, however, as more advanced mathematics texts (as well as the Web site Wolfram|Alpha) use \(\log\) for the (base \(e\)) natural logarithm.}

Example \(\PageIndex{2}\)

Use the Change of Base formula and your calculator to find \(\log_{\pi}(7)\) and \(\log_2(8)\).

Solution

\(\displaystyle \log_{\pi}(7) = \frac{\ln(7)}{\ln(\pi)} \approx \frac{1.946}{1.145} \approx 1.700\). (Check that \(\pi^{1.7} \approx 7\).) Likewise, \(\displaystyle \log_2(8) = \frac{\ln(8)}{\ln(2)} = 3\).

Practice \(\PageIndex{1}\)

Find the values of \(\log_9 20\), \(\log_3 20\) and \(\log_{\pi} e\).

Answer

\(\displaystyle \log_9(20) = \frac{\log(20)}{\log(9)} \approx 1.3634165 \approx \frac{\ln( 20 )}{\ln( 9 )}\)

\(\displaystyle \log_3(20) = \frac{\log(20)}{\log(3)} \approx 2.726833 \approx \frac{\ln( 20 )}{\ln( 3 )}\)

\(\displaystyle \log_{\pi}(e) = \frac{\log(e)}{\log(\pi)} \approx 0.8735685 \approx \frac{\ln(e)}{\ln(\pi)} = \frac{1}{\ln(\pi)}\)

Putting \(b = e\) in the Change of Base Formula, \(\displaystyle \log_a(x) = \frac{\log_e(x)}{\log_e(a)} = \frac{\ln(x)}{\ln(a)}\), so any logarithm can be written as a natural logarithm divided by a constant. This makes any logarithmic function easy to differentiate.

Theorem

\(\displaystyle \mbox{D}\left( \log_a( x ) \right) = \frac{1}{x \ln(a)}\) and \(\displaystyle \mbox{D}\left( \log_a ( f(x) ) \right) = \frac{f'(x)}{f(x)}\cdot \frac{1}{\ln(a)}\)

Proof

\(\displaystyle \mbox{D}\left( \log_a( x ) \right) = \mbox{D}\left(\frac{\ln x}{\ln a}\right) = \frac{1}{\ln(a)} \cdot \mbox{D}( \ln x ) = \frac{1}{\ln(a)}\cdot \frac{1}{x} = \frac{1}{x \ln(a)}\).  The second differentiation formula follows from the Chain Rule.

Practice \(\PageIndex{2}\)

Calculate \(\mbox{D}\left( \log_{10} ( \sin(x) ) \right)\) and \(\mbox{D}\left( \log_{\pi} ( e^x ) \right)\).

Answer

\(\displaystyle \mbox{D}\left( \log_{10} ( \sin(x) ) \right) =  \frac{1}{\sin(x) \cdot \ln(10)} \mbox{D}( \sin(x) ) = \frac{\cos(x)}{\sin(x)\cdot \ln(10)}\)

\(\displaystyle \mbox{D}\left(\log_{\pi}( e^x ) \right) = \frac{1}{e^x \cdot \ln(\pi)} \mbox{D}( e^x) = \frac{e^x}{e^x \cdot \ln(\pi)} = \frac{1}{\ln(\pi)}\)

The number \(e\) might seem like an “unnatural” base for a natural logarithm, but of all the possible bases, the logarithm with base \(e\) has the nicest and easiest derivative. The natural logarithm is even related to the distribution of prime numbers. In 1896, the mathematicians Hadamard and Vallèe-Poussin proved the following conjecture of Gauss (the Prime Number Theorem): For large values of \(N\),\[\mbox{number of primes less than }N \approx \frac{N}{\ln(N)}\nonumber\]

Derivative of \(a^x\)

Once we know the derivative of \(e^x\) and the Chain Rule, it is relatively easy to determine the derivative of \(a^x\) for any \(a > 0\).

Theorem

\(\mbox{D}( a^x) = a^x \cdot \ln(a)\) \ for \(a > 0\).

Proof

If \(a > 0\), then \(a^x > 0\) and \(\displaystyle a^x = e^{\ln( a^x )} = e^{x \cdot \ln(a)}\), so we have: \(\displaystyle \mbox{D}( a^x) = \mbox{D}\left ( e^{\ln( a^x )} \right) = \mbox{D}\left( e^{x\cdot \ln(a)} \right) = e^{x\cdot \ln(a)}\cdot \mbox{D}( x\cdot \ln(a)) = a^x \cdot \ln(a)\).

Example \(\PageIndex{3}\)

Calculate \(\displaystyle \mbox{D}( 7^x)\) and \(\displaystyle \frac{d}{dt} \left( 2^{\sin(t)}\right)\).

Solution

\(\displaystyle \mbox{D}( 7^x) = 7^x \cdot \ln(7) \approx (1.95) 7^x\). We can write \(\displaystyle y = 2^{\sin(t)}\) as \(y = 2^u\) with \(u = \sin(t)\). Using the Chain Rule: \(\displaystyle \frac{dy}{dt} =  \frac{dy}{du}\cdot \frac{du}{dt} = 2^u \cdot \ln(2) \cos(t) = 2^{\sin(t)} \cdot \ln(2) \cdot \cos(t)\).

Practice \(\PageIndex{3}\)

Calculate \(\displaystyle \mbox{D}\left( \sin( 2^x ) \right)\) and \(\displaystyle \frac{d}{dt}\left( 3^{t^2}\right)\).

Answer

\(\displaystyle \mbox{D}\left( \sin( 2^x ) \right) = \cos( 2^x ) \mbox{D}\left( 2^x \right) = \cos( 2^x ) \cdot 2^x \cdot \ln(2)\)

\(\displaystyle \frac{d}{dt}\left(3^{t^2}\right) = 3^{t^2} \ln(3) \mbox{D}(t^2) = 3^{t^2} \ln(3) \cdot 2t\) 

Some Applied Problems

Let’s examine some applications involving more complicated functions.

Example \(\PageIndex{4}\)

A ball at the end of a rubber band is oscillating up and down, and its height (in feet) above the floor at time \(t\) seconds is \(\displaystyle h(t) = 5 + 2\sin\left( \frac{t}{2} \right)\) (with \(t\) in radians).

1. How fast is the ball traveling after 2 seconds? After 4 seconds? After 60 seconds?
2. Is the ball moving up or down after 2 seconds? After 4 seconds? After 60 seconds?
3. Is the vertical velocity of the ball ever \(0\)?

Solution

1. \(\displaystyle v(t) = h'(t) = \mbox{D}\left( 5 + 2\sin\left( \frac{t}{2} \right)\right) = 2\cos\left( \frac{t}{2} \right) \cdot \frac12\) so \(v(t) = \cos\left( \frac{t}{2} \right)\) feet/second: \(v( 2 ) = \cos( \frac22 ) \approx 0.540\) ft/s, \(v( 4 ) = \cos( \frac42 ) \approx -0.416\) ft/s, and \(v( 60 ) = \cos( \frac{60}{2} ) \approx 0.154\) ft/s.
2. The ball is moving up at \(t = 2\) and \(t = 60\), down when \(t = 4\).
3. \(\displaystyle v( t ) = \cos\left( \frac{t}{2} \right) = 0\) when \(\displaystyle \frac{t}{2} = \frac{\pi}{2} \pm k\cdot \pi \Rightarrow t = \pi \pm 2\pi k\) for integer \(k\).

Example \(\PageIndex{5}\)

If 2,400 people now have a disease, and the number of people with the disease appears to double every 3 years, then the number of people expected to have the disease in \(t\) years is \(\displaystyle y = 2400\cdot 2^{\frac{t}{3}}\).

1. How many people are expected to have the disease in 2 years?
2. When are 50,000 people expected to have the disease?
3. How fast is the number of people with the disease growing now? How fast is it expected to be growing 2 years from now?

Solution

1. In 2 years, \(\displaystyle y = 2400\cdot 2^{\frac23} \approx\) 3,810 people.
2. We know \(y = 50000\) and need to solve \(\displaystyle 50000 = 2400\cdot 2^{\frac{t}{3}}\) for \(t\). Taking logarithms of each side of the equation: \(\displaystyle \ln(50000) = \ln\left(2400\cdot 2^{\frac23}\right) = \ln(2400) + \frac{t}{3}\cdot\ln(2)\) so \(10.819 \approx 7.783 + 0.231t\) and \(t \approx 13.14\) years. We expect 50,000 people to have the disease about 13 years from now.
3. This question asks for \(\displaystyle \frac{dy}{dt}\) when \(t = 0\) and \(t = 2\).\[\frac{dy}{dt} = \frac{d}{dt}\left( 2400\cdot 2^{\frac{t}{3}}\right) = 2400\cdot 2^{\frac{t}{3}}\cdot\ln(2)\cdot\frac13 \approx 554.5 \cdot 2^{\frac{t}{3}}\nonumber\]Now, at \(t = 0\), the rate of growth of the disease is approximately \hspace*{1.5em}\)554.5\cdot 2^0 \approx 554.5\) people/year. In 2 years, the rate of growth will be approximately \(554.5\cdot 2^{\frac23} \approx 880\) people/year.

Example \(\PageIndex{6}\)

You are riding in a balloon, and at time \(t\) (in minutes) you are \(h(t) = t + \sin(t)\) thousand feet above sea level. If the temperature at an elevation \(h\) is \(\displaystyle T(h) = \frac{72}{1 + h}\) degrees Fahrenheit, then how fast is the temperature changing when \(t = 5\) minutes?

Solution

As \(t\) changes, your elevation will change. And, as your elevation changes, so will the temperature. It is not difficult to write the temperature as a function of time, and then we could calculate \(\displaystyle \frac{dT}{dt} = T'(t)\) and evaluate \(T'(5)\). Or we could use the Chain Rule:\[\frac{dT}{dt} = \frac{dT}{dh} \cdot \frac{dh}{dt} = -\frac{72}{(1+h)^2} \cdot (1+\cos(t))\nonumber\]At \(t = 5\), \(h(5) = 5 + \sin(5) \approx 4.04\) so \(T'(5) \approx -\frac{72}{(1 + 4.04)^2} \cdot (1 + 0.284) \approx -3.64\) \(^{\circ}\)/minute.

Practice \(\PageIndex{4}\)

Write the temperature \(T\) in the previous example as a function of the variable \(t\) alone and then differentiate \(T\) to determine the value of \(\displaystyle \frac{dT}{dt}\) when \(t = 5\) minutes.

Answer

\(\displaystyle T =  \frac{72}{1 + h} =  \frac{72}{1 + t + \sin(t)} \Rightarrow\)

\(\displaystyle \frac{dT}{dt} = \frac{(1+t+\sin(t))\cdot 0 - 72\cdot\mbox{D}(1+t+\sin(t))}{(1+t+\sin(t))^2} = \frac{-72(1+\cos(t))}{(1+t+\sin(t))^2}\)

When \(t = 5\), \(\displaystyle \frac{dT}{dt} = \frac{-72(1+\cos(5))}{(1+5+\sin(5))^2} \approx -3.63695\).

Example \(\PageIndex{7}\)

A scientist has determined that, under optimum conditions, an initial population of 40 bacteria will grow “exponentially” to \(\displaystyle f(t) = 40\cdot e^{\frac{t}{5}}\) bacteria after \(t\) hours.

1. Graph \(y = f(t)\) for \(0 \leq t \leq 15\). Calculate \(f(0)\), \(f(5)\) and \(f(10)\).
2. How fast is the population increasing at time \(t\)? (Find \(f'(t)\).)
3. Show that the rate of population increase, \(f'(t)\), is proportional to the population, \(f(t)\), at any time \(t\). (Show \(f'(t) = K\cdot f(t)\) for some constant \(K\).)

Solution

1. The graph of \(y = f(t)\) appears below:
   
   \(f(0) = 40\cdot e^{\frac05} = 40\) bacteria, \(f(5) = 40\cdot e^{\frac55} = 40e \approx 109\) bacteria and \(f(10) = 40\cdot e^{\frac{10}{5}} \approx 296\) bacteria.
2. \(f'(t) = \frac{d}{dt} \left( f(t) \right) = \frac{d}{dt} \left( 40\cdot e^{\frac{t}{5}}\right) = 40\cdot e^{\frac{t}{5}} \cdot \frac{d}{dt}\left( \frac{t}{5}\right) = 40\cdot e^{\frac{t}{5}} \cdot \frac15 = 8\cdot e^{\frac{t}{5}}\) bacteria/hour.
3. \(f'(t) = 8\cdot e^{\frac{t}{5}} = \frac15 \cdot 40 e^{\frac{t}{5}} = \frac15 f(t)\) so \(f'(t) = K\cdot f(t)\) with \(K = \frac15\). The rate of change of the population is proportional to its size.

Parametric Equations

Suppose a robot has been programmed to move in the \(xy\)-plane so at time \(t\) its \(x\)-coordinate will be \(\sin(t)\) and its \(y\)-coordinate will be \(t^2\). Both \(x\) and \(y\) are functions of the independent parameter \(t\):  \(x(t) = \sin(t)\) and \(y(t) = t^2\). The path of the robot can be found by plotting \((x,y) = \left( x(t), y(t) \right)\) for lots of values of \(t\):

\(t\)	\(x(t) = \sin(t)\)	\(y(t) = t^2\)	point
\(0\)	\(0\)	\(0\)	\((0, 0)\)
\(0.5\)	\(0.48\)	\(0.25\)	\((0.48, 0.25)\)
\(1.0\)	\(0.84\)	\(1\)	\((0.84, 1)\)
\(1.5\)	\(1.00\)	\(2.25\)	\((1, 2.25)\)
\(2.0\)	\(0.91\)	\(4\)	\((0.91, 4)\)

Typically we know \(x(t)\) and \(y(t)\) and need to find \(\displaystyle \frac{dy}{dx}\), the slope of the tangent line to the graph of \(\left( x(t), y(t) \right)\). The Chain Rule says:\[\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}\nonumber\]so , algebraically solving for \(\displaystyle \frac{dy}{dx}\), we get:\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\nonumber\]If we can calculate \(\displaystyle \frac{dy}{dt}\) and \(\displaystyle \frac{dx}{dt}\), the derivatives of \(y\) and \(x\) with respect to the parameter \(t\), then we can determine \(\displaystyle \frac{dy}{dx}\), the rate of change of \(y\) with respect to \(x\).

Theorem

If : \(x = x(t)\) and \(y = y(t)\) are differentiable with respect to \(t\) and \(\displaystyle \frac{dx}{dt} \neq 0\)
     
then: \(\displaystyle \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\).

Example \(\PageIndex{8}\)

Find the slope of the tangent line to the graph of \((x,y) = \left( \sin(t), t^2 \right)\) when \(t = 2\).

Solution

\(\displaystyle \frac{dx}{dt} = \cos(t)\) and \(\displaystyle \frac{dy}{dt} = 2t\). When \(t = 2\), the object is at the point \((\sin(2), 2^2) \approx (0.91,4 )\) and the slope of the tangent line is:\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t}{\cos(t)} = \frac{2\cdot 2}{\cos(2)} \approx \frac{4}{-0.42} \approx -9.61\nonumber\]Notice in the figure that the slope of the tangent line to the curve at \((0.91,4 )\) is negative and very steep.

Practice \(\PageIndex{5}\)

Graph \((x,y) = \left( 3\cos(t), 2\sin(t) \right)\) and find the slope of the tangent line when \(t = \frac{\pi}{2}\).

Answer

\(\displaystyle x(t) = 3\cos(t) \Rightarrow \frac{dx}{dt} = -3\sin(t)\), \(y(t) = 2\sin(t) \Rightarrow \frac{dy}{dt} = 2\cos(t)\):\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2\cos(t)}{-3\sin(t)} \Rightarrow \left.\frac{dy}{dx}\right|_{t = \frac{\pi}{2}} = \frac{2\cos(\frac{\pi}{2})}{-3\sin(\frac{\pi}{2})} = \frac{2\cdot 0}{-3\cdot 1} = 0\nonumber\]

When we calculated \(\displaystyle \frac{dy}{dx}\), the slope of the tangent line to the graph of \(( x(t), y(t) )\), we used the derivatives \(\displaystyle \frac{dx}{dt}\) and \(\displaystyle \frac{dy}{dt}\). Each of these also has a geometric meaning: \(\displaystyle \frac{dx}{dt}\) measures the rate of change of \(x(t)\) with respect to \(t\): it tells us whether the \(x\)-coordinate is increasing or decreasing as the \(t\)-variable increases (and how fast it is changing), while \(\displaystyle \frac{dy}{dt}\) measures the rate of change of \(y(t)\) with respect to \(t\).

Example \(\PageIndex{9}\)

For the parametric graph below:

determine whether \(\displaystyle \frac{dx}{dt}\), \(\displaystyle \frac{dy}{dt}\) and \(\displaystyle \frac{dy}{dx}\) are positive or negative when \(t=2\).

Solution

As we move through the point \(B\) (where \(t=2\)) in the direction of increasing values of \(t\), we are moving to the left, so \(x(t)\) is decreasing and \(\displaystyle \frac{dx}{dt} < 0\). The values of \(y(t)\) are increasing, so \(\displaystyle \frac{dy}{dt} > 0\). Finally, the slope of the tangent line, \(\displaystyle \frac{dy}{dx}\), is negative.

As a check on the sign of \(\displaystyle \frac{dy}{dx}\) in the previous example:\[\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} =\frac{\mbox{positive}}{\mbox{negative}} = \mbox{negative}\nonumber\]

Practice \(\PageIndex{6}\)

For the parametric graph in Example \(\PageIndex{9}\), tell whether \(\displaystyle \frac{dx}{dt}\), \(\displaystyle \frac{dy}{dt}\) and \(\displaystyle \frac{dy}{dx}\) are positive or negative at \(t=1\) and \(t=3\).

Answer

\(x=1\): positive, positive, positive. \(x=3\): positive, negative, negative.

Speed

If we know the position of an object at any time, then we can determine its speed. The formula for speed comes from the distance formula and looks a lot like it, but involves derivatives.

Theorem

If: \(x = x(t)\) and \(y = y(t)\) give the location of an object at time \(t\) and both are differentiable functions of \(t\)

then: the speed of the object is\[\sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\nonumber\]

Proof

The speed of an object is the limit, as \(\Delta t \rightarrow 0\), of:
\begin{align*}\frac{\mbox{change in position}}{\mbox{change in time}} & = \frac{\sqrt{(\Delta x)^2 + (\Delta y)^2}}{\Delta t} = \sqrt{\frac{(\Delta x)^2 + (\Delta y)^2}{(\Delta t)^2}}\\
& = \sqrt{\left(\frac{\Delta x}{\Delta t}\right)^2 + \left(\frac{\Delta y}{\Delta t}\right)^2} \rightarrow \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\end{align*}
as \(\Delta t \rightarrow 0\).

Example \(\PageIndex{10}\)

Find the speed of the object whose location at time \(t\) is \((x,y) = \left( \sin(t), t^2 \right)\) when \(t = 0\) and \(t = 1\).

Solution

\(\displaystyle \frac{dx}{dt} = \cos(t)\) and \(\displaystyle \frac{dy}{dt} = 2t\) so:\[\mbox{speed } = \sqrt{\left(\cos(t)\right)^2 + (2t)^2} = \sqrt{\cos^2(t) + 4t^2}]When \(t = 0\), speed \(= \sqrt{\cos^2(0) + 4(0)^2} = \sqrt{1 + 0} = 1\). When \(t = 1\), speed \(= \sqrt{\cos^2(1) + 4(1)^2} \approx \sqrt{0.29 + 4} \approx 2.07\).

Practice \(\PageIndex{7}\)

Show that an object located at \((x,y) = \left( 3\sin(t), 3\cos(t) \right)\) at time \(t\) has a constant speed. (This object is moving on a circular path.)

Answer

\(x(t) = 3\sin(t) \Rightarrow \frac{dx}{dt} = 3\cos(t)\) and \(y(t) = 3\cos(t) \Rightarrow \frac{dy}{dt} = -3\sin(t)\). So:
\begin{align*}\mbox{speed} &= \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \sqrt{(3\cos(t))^2 + (-3\sin(t))^2}\\
&= \sqrt{9\cdot \cos^2(t) + 9\cdot \sin^2(t)} = \sqrt{9} = 3 \quad \mbox{(a constant)}\end{align*}

Practice \(\PageIndex{8}\)

Is the object at \((x,y) = \left( 3\cos(t), 2\sin(t) \right)\) at time \(t\) traveling faster at the top of the ellipse (\)t = \frac{\pi}{2}\)) or at the right edge (\(t = 0\))?

Answer

\(x(t) = 3\cos(t) \Rightarrow \frac{dx}{dt} = -3\sin(t)\) and \(y(t) = 2\sin(t) \Rightarrow \frac{dy}{dt} = 2\cos(t)\) so:
\begin{align*}\mbox{speed} &= \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} = \sqrt{(-3\sin(t))^2 + (2\cos(t))^2}\\ 
&= \sqrt{9\cdot \sin^2(t) + 4\cdot \cos^2(t)}\end{align*}
When \(t = 0\), the speed is \(\sqrt{9\cdot 0^2 + 4\cdot 1^2} = 2\).

When \(t = \frac{\pi}{2}\), the speed is \(\sqrt{9\cdot 1^2 + 4\cdot 0^2} = 3\) (faster).

Problems

In Problems 1–27, differentiate the given function.

1. \(\ln( 5x )\)
2. \(\ln( x^2 )\)
3. \(\ln( x^k )\)
4. \(\ln( x^x ) = x\cdot \ln(x)\)
5. \(\ln( \cos(x) )\)
6. \(\cos( \ln(x) )\)
7. \(\log_2(5x)\)
8. \(\log_2(kx)\)
9. \(\ln( \sin(x) )\)
10. \(\ln( kx )\)
11. \(\log_2( \sin(x) )\)
12. \(\ln( e^x )\)
13. \(\log_5(5^x)\)
14. \(\ln\left( e^{f(x)} \right)\)
15. \(x\cdot \ln( 3x )\)
16. \(e^x \cdot \ln(x)\)
17. \(\displaystyle \frac{\ln(x)}{x}\)
18. \(\displaystyle \sqrt{x + \ln(3x)}\)
19. \(\ln\left(\sqrt{5x-3}\right)\)
20. \(\ln( \cos(t) )\)
21. \(\cos( \ln( w ) )\)
22. \(\ln( ax + b )\)
23. \(\ln\left(\sqrt{t+1}\right)\)
24. \(3^x\)
25. \(5^{\sin(x)}\)
26. \(x \cdot \ln(x) - x\)
27. \(\ln\left( \sec(x) + \tan(x) \right)\)
28. Find the slope of the line tangent to \(f(x) = \ln( x )\) at the point \(( e , 1 )\). Find the slope of the line tangent to \(g(x) = e^x\) at the point \(( 1, e)\). How are the slopes of \(f\) and \(g\) at these points related?
29. Find a point \(P\) on the graph of \(f(x) = \ln(x)\) so the tangent line to \(f\) at \(P\) goes through the origin.
30. You are moving from left to right along the graph of \(y = \ln( x )\):
    
    1. If the \(x\)-coordinate of your location at time \(t\) seconds is \(x(t) = 3t +2\), then how fast is your elevation increasing?
    2. If the \(x\)-coordinate of your location at time \(t\) seconds is \(x(t) = e^t\), then how fast is your elevation increasing?
31. The percent of a population, \(p(t)\), who have heard a rumor by time \(t\) is often modeled by \(\displaystyle p(t) = \frac{100}{1 + Ae^{-t}} = 100\left(1 + Ae^{-t}\right)^{-1}\) for some positive constant \(A\). Calculate \(p'(t)\), the rate at which the rumor is spreading.
32. If we start with \(A\) atoms of a radioactive material that has a “half-life” (the time it takes for half of the material to decay) of 500 years, then the number of radioactive atoms left after \(t\) years is \(\displaystyle r(t) = A\cdot e^{-Kt}\) where \(\displaystyle K = \frac{\ln(2)}{500}\). Calculate \(r'(t)\) and show that \(r'(t)\) is proportional to \(r(t)\) (that is, \(r'(t) = b\cdot r(t)\) for some constant \(b\)).

In Problems 33–41, find a function with the given derivative.

33. \(\displaystyle f'(x) = \frac{8}{x}\)
34. \(\displaystyle h'(x) = \frac{3}{3x + 5}\)
35. \(\displaystyle f'(x) = \frac{\cos(x)}{3 + \sin(x)}\)
36. \(\displaystyle g'(x) = \frac{x}{1 + x^2}\)
37. \(\displaystyle g'(x) = 3e^{5x}\)
38. \(\displaystyle h'(x) = e^2\)
39. \(\displaystyle f'(x) = 2x\cdot e^{x^2}\)
40. \mbox{\(\displaystyle g'(x) = \cos(x) e^{\sin(x)}\)}
41. \(\displaystyle h'(x) = \cot(x) = \frac{\cos(x)}{\sin(x)}\)
42. Define \(A(x)\) to be the area bounded between the \(t\)-axis, the graph of \(y = f(t)\) and a vertical line at \(t = x\):
    
    The area under each “hump” of \(f\) is \(2\) square inches.
    1. Graph \(A(x)\) for \(0 \leq x \leq 9\).
    2. Graph \(A'(x)\) for \(0 \leq x \leq 9\).

Problems 43--48 involve parametric equations.

43. At time \(t\) minutes, robot A is at \(( t, 2t + 1)\) and robot \(B\) is at \(( t^2, 2t^2 + 1)\).
    1. Where is each robot when \(t=0\) and \(t = 1\)?
    2. Sketch the path each robot follows during the first minute.
    3. Find the slope of the tangent line, \(\displaystyle \frac{dy}{dx}\), to the path of each robot at \(t = 1\) minute.
    4. Find the speed of each robot at \(t = 1\) minute.
    5. Discuss the motion of a robot that follows the path \((\sin(t), 2\sin(t) + 1 )\) for \(20\) minutes.
44. Let \(x(t) = t + 1\) and \(y(t) = t^2\).
    1. Graph \(( x(t), y(t) )\) for \(-1 \leq t \leq 4\).
    2. Find \(\displaystyle \frac{dx}{dt}\), \(\displaystyle \frac{dy}{dt}\), the tangent slope \(\displaystyle \frac{dy}{dx}\), and speed when \(t = 1\) and \(t = 4\).
45. For the parametric graph shown below:
    
    determine whether \(\displaystyle \frac{dx}{dt}\), \(\displaystyle \frac{dy}{dt}\) and \(\displaystyle \frac{dy}{dx}\) are positive, negative or \(0\) when \(t = 1\) and \(t=3\).
46. For the parametric graph shown below:
    
    determine whether \(\displaystyle \frac{dx}{dt}\), \(\displaystyle \frac{dy}{dt}\) and \(\displaystyle \frac{dy}{dx}\) are positive, negative or \(0\) when \(t = 1\) and \(t=3\).
47. The parametric graph \(( x(t), y(t) )\) defined by \(x(t) = R\cdot (t - \sin(t) )\) and \(y(t) = R\cdot (1 - \cos(t) )\) is called a cycloid, the path of a light attached to the edge of a rolling wheel with radius \(R\).
    1. Graph \(( x(t), y(t) )\) for \(0 \leq t \leq 4\pi\).
    2. Find \(\displaystyle \frac{dx}{dt}\), \(\displaystyle \frac{dy}{dt}\), the tangent slope \(\displaystyle \frac{dy}{dx}\), and speed when \(t = \frac{\pi}{2}\) and \(t = \pi\).
48. Describe the motion of particles whose locations at time \(t\) are \(( \cos(t), \sin(t) )\) and \(( \cos(t), -\sin(t) )\).
49. 1. Describe the path of a robot whose location at time \(t\) is \(( 3\cdot \cos(t), 5\cdot \sin(t) )\).
    2. Describe the path of a robot whose location at time \(t\) is \(( A\cdot \cos(t), B\cdot \sin(t) )\).
    3. Give parametric equations so the robot will move along the same path as in part (a) but in the opposite direction.
50. After \(t\) seconds, a projectile hurled with initial velocity \(v\) and angle \(\theta\) will be at \(x(t) = v\cdot \cos(\theta) \cdot t\) feet and \(y(t) = v\cdot\sin(\theta)\cdot t - 16t^2\) feet:
    
    (This formula neglects air resistance.)
    1. For an initial velocity of \(80\) feet/second and an angle of \(\frac{\pi}{4}\), find \(T > 0\) so that \(y(T) = 0\). What does this value for \(t\) represent physically? Evaluate \(x(T)\).
    2. For \(v\) and \(\theta\) in part (a), calculate \(\displaystyle \frac{dy}{dx}\). Find \(T\) so that \(\displaystyle \frac{dy}{dx} = 0\) at \(t = T\), and evaluate \(x(T)\). What does \(x(T)\) represent physically?
    3. What initial velocity is needed so a ball hit at an angle of \(\frac{\pi}{4} \approx 0.7854\) will go over a 40-foot-high fence 350 feet away?
    4. What initial velocity is needed so a ball hit at an angle of \(0.7\) radians will go over a 40-foot-high fence 350 feet away?
51. Use the method from the proof that \(\mbox{D}(\ln(x)) = \frac{1}{x}\) to compute the derivative \(\mbox{D}(\arctan(x))\):
    1. Rewrite \(y = \arctan(x)\) as \(\tan(y) = x\).
    2. Differentiate both sides using the Chain Rule and solve for \(y'\).
    3. Use the identity \(1 + \tan^2(\theta) = \sec^2(\theta)\) and the fact that \(\tan(y) = x\) to show that \(\displaystyle y' = \frac{1}{1+x^2}\).
52. Use the method from the proof that \(\mbox{D}(\ln(x)) = \frac{1}{x}\) to compute the derivative \(\mbox{D}(\arcsin(x))\):
    1. Rewrite \(y = \arcsin(x)\) as \(\sin(y) = x\).
    2. Differentiate both sides using the Chain Rule and solve for \(y'\).
    3. Use the identity \(\cos^2(\theta) + \sin^2(\theta) = 1\) and the fact that \(\sin(y) = x\) to show that \(\displaystyle y' = \frac{1}{\sqrt{1-x^2}}\).