Activity 26.
Use the product rule to respond to each of the prompts below. Throughout, be sure to carefully label any derivative you find by name. It is not necessary to algebraically simplify any of the derivatives you compute.
(a)
Let \(m(w)=3w^{17} 4^w\text{.}\) Find \(m'(w)\text{.}\)
Hint.
Let the first function be \(3w^{17}\text{.}\)
Answer.
\(m'(w) = 3w^{17} \cdot 4^w \ln(4) + 4^w \cdot 51w^{16}\text{.}\)
Solution.
By the product rule,
\begin{equation*}
m'(w) = 3w^{17} \cdot 4^w \ln(4) + 4^w \cdot 51w^{16}\text{.}
\end{equation*}
(b)
Let \(h(t) = (\sin(t) + \cos(t))t^4\text{.}\) Find \(h'(t)\text{.}\)
Hint.
Let the first function be \((\sin(t) + \cos(t))\text{.}\)
Answer.
\(h'(t) = (\sin(t) + \cos(t)) \cdot 4t^3 + t^4 \cdot (\cos(t) - \sin(t))\text{.}\)
Solution.
By the product rule,
\begin{equation*}
h'(t) = (\sin(t) + \cos(t)) \cdot 4t^3 + t^4 \cdot (\cos(t) - \sin(t))\text{.}
\end{equation*}
(c)
Determine the slope of the tangent line to the curve \(y = f(x)\) at the point \((1, f(1))\) if \(f\) is given by the rule \(f(x) = e^x \sin(x)\text{.}\)
Hint.
Remember that the slope of the tangent line to \(y = f(x)\) at \((a,f(a))\) is given by \(f'(a)\text{.}\)
Answer.
\(f'(1) = e(\cos(1) + \sin(1)) \approx 3.756\text{.}\)
Solution.
To determine the slope of the tangent line at \((1, f(1))\text{,}\) we want to find \(f'(1)\text{.}\) Since \(f(x) = e^x \sin(x)\text{,}\) the product rule tells us that \(f'(x) = e^x \cdot \cos(x) + \sin(x) \cdot e^x\text{.}\) Thus, \(f'(1) = e^1 \cdot \cos(1) + \sin(1) \cdot e^1 = e(\cos(1) + \sin(1)) \approx 3.756\text{.}\)
(d)
Find the tangent line approximation \(L(x)\) to the function \(y = g(x)\) at the point \((-1,g(-1))\) if \(g\) is given by the rule \(g(x) = (x^2 + x) 2^x\text{.}\)
Hint.
Remember that \(L(x) = g(a) + g'(a)(x-a)\text{.}\)
Answer.
\(L(x) = -\frac{1}{2}(x+1)\text{.}\)
Solution.
First, observe that \(g(-1) = ((-1)^2 - 1) \cdot 2^{-1} = 0\text{.}\) Further, by the product rule, \(g'(x) = (x^2 + x) \cdot 2^x \ln(2) + 2^x \cdot (2x + 1)\text{,}\) and therefore \(g'(-1) = ((-1)^2 - 1) \cdot 2^{-1} \ln(2) + 2^{-1} \cdot (2(-1) + 1) = 0 + \frac{1}{2}(-1) = -\frac{1}{2}\text{.}\) Therefore,
\begin{equation*}
L(x) = g(-1) + g'(-1)(x+1) = 0 - \frac{1}{2}(x+1) = -\frac{1}{2}(x+1)\text{.}
\end{equation*}
