Activity 27.
Use relevant derivative rules to respond to each of the prompts below. Throughout, be sure to use proper notation and carefully label any derivative you find by name.
(a)
Let \(f(r) = (5r^3 + \sin(r))(4^r - 2\cos(r))\text{.}\) Find \(f'(r)\text{.}\)
Hint.
Observe that \(f\) is fundamentally a product. Which is the first function? The second?
Answer.
\(f'(r) = (5r^3 + \sin(r))[4^r \ln(4) + 2\sin(r)] + (4^r - 2\cos(r))[15r^2 + \cos(r)]\text{.}\)
Solution.
Using the product rule, followed by the sum and constant multiple rule, observe that
\begin{align*}
f'(r) =\mathstrut \amp (5r^3 + \sin(r))\frac{d}{dr}[4^r - 2\cos(r)] + (4^r - 2\cos(r))\frac{d}{dr}[5r^3 + \sin(r)]\\
=\mathstrut \amp (5r^3 + \sin(r))[4^r \ln(4) + 2\sin(r)] + (4^r - 2\cos(r))[15r^2 + \cos(r)]
\end{align*}
(b)
Let \(\displaystyle p(t) = \frac{\cos(t)}{t^6 \cdot 6^t}\text{.}\) Find \(p'(t)\text{.}\)
Hint.
Note that \(p\) has the overall structure of a quotient.
Answer.
\(p'(t) = \frac{t^6 \cdot 6^t [-\sin(t)] - \cos(t) [t^6 \cdot 6^t \ln(6) + 6^t \cdot 6t^5]}{(t^6 \cdot 6^t)^2}\text{.}\)
Solution.
We use the quotient rule on \(p\text{,}\) followed by the product rule to differentiate the denominator, finding that
\begin{align*}
p'(t) =\mathstrut \amp \frac{t^6 \cdot 6^t \frac{d}{dt}[\cos(t)] - \cos(t) \frac{d}{dt}[t^6 \cdot 6^t]}{(t^6 \cdot 6^t)^2}\\
=\mathstrut \amp \frac{t^6 \cdot 6^t [-\sin(t)] - \cos(t) [t^6 \cdot 6^t \ln(6) + 6^t \cdot 6t^5]}{(t^6 \cdot 6^t)^2}
\end{align*}
(c)
Let \(g(z) = 3z^7 e^z - 2z^2 \sin(z) + \frac{z}{z^2 + 1}\text{.}\) Find \(g'(z)\text{.}\)
Hint.
Think about how \(g\) is a sum of three functions. What is the structure of each of the three functions in the sum?
Answer.
\(g'(z) = 3 [z^7 e^z + 7z^6e^z] - 2[z^2 \cos(z) + 2z\sin(z)] + \frac{(z^2+1) 1 - z(2z)}{(z^2 + 1)^2}\text{.}\)
Solution.
Using the sum and constant multiple rules, it follows first that
\begin{equation*}
g'(z) = 3 \frac{d}{dz}[z^7 e^z] - 2\frac{d}{dz}[z^2 \sin(z)] + \frac{d}{dz}\left[ \frac{z}{z^2 + 1} \right]\text{.}
\end{equation*}
Applying the product rule in the first two terms and the quotient rule in the third, we find that
\begin{equation*}
g'(z) = 3 [z^7 e^z + 7z^6e^z] - 2[z^2 \cos(z) + 2z\sin(z)] + \frac{(z^2+1) 1 - z(2z)}{(z^2 + 1)^2}\text{.}
\end{equation*}
(d)
A moving particle has its position in feet at time \(t\) in seconds given by the function \(s(t) = \frac{3\cos(t) - \sin(t)}{e^t}\text{.}\) Find the particle’s instantaneous velocity at the moment \(t = 1\text{.}\)
Hint.
How is the velocity of a moving object related to its position?
Answer.
\(s'(1) = \frac{-2\sin(1)-4\cos(1)}{e^1} \approx -1.414\) feet per second.
Solution.
The particle’s instantaneous velocity at the moment \(t = 1\) is given by \(s'(1)\text{.}\) We use the quotient rule to find \(s'(t)\text{,}\) and simplify by removing a common factor of \(e^t\) to get
\begin{equation*}
s'(t) = \frac{e^t(-3\sin(t) - \cos(t))-(3\cos(t) - \sin(t))e^t}{(e^t)^2} = \frac{-2\sin(t)-4\cos(t)}{e^t}\text{.}
\end{equation*}
Thus, \(s'(1) = \frac{-2\sin(1)-4\cos(1)}{e^1} \approx -1.414\text{,}\) which is the particle’s instantaneous velocity in feet per second at the moment \(t = 1\text{.}\)
(e)
Suppose that \(f(x)\) and \(g(x)\) are differentiable functions and it is known that \(f(3) = -2\text{,}\) \(f'(3) = 7\text{,}\) \(g(3) = 4\text{,}\) and \(g'(3) = -1\text{.}\) If \(p(x) = f(x) \cdot g(x)\) and \(\displaystyle q(x) = \frac{f(x)}{g(x)}\text{,}\) calculate \(p'(3)\) and \(q'(3)\text{.}\)
Hint.
Since we know \(p(x) = f(x) \cdot g(x)\text{,}\) it follows \(p'(x) = f(x) g'(x) + g(x) f'(x)\text{.}\)
Answer.
\(p'(3) = 30\) and \(q'(3) = \frac{13}{8}\text{.}\)
Solution.
Since \(p(x) = f(x) \cdot g(x)\text{,}\) the product rule tells us
\begin{equation*}
p'(x) = f(x)g'(x) + g(x)f'(x)\text{,}
\end{equation*}
and since \(\displaystyle q(x) = \frac{f(x)}{g(x)}\text{,}\) by the quotient rule we know
\begin{equation*}
q'(x) = \frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}\text{.}
\end{equation*}
Using the given information (\(f(3) = -2\text{,}\) \(f'(3) = 7\text{,}\) \(g(3) = 4\text{,}\) and \(g'(3) = -1\)) we now see that
\begin{equation*}
p'(3) = f(3)g'(3) + g(3)f'(3) = (-2)(-1) + (4)(7) = 30
\end{equation*}
and
\begin{equation*}
q'(3) = \frac{g(3)f'(3)-f(3)g'(3)}{g(3)^2} = \frac{(4)(7) - (-2)(-1)}{4^2} = \frac{13}{8}\text{.}
\end{equation*}
