Question

7 fill in the blank 3 points
the graphs of f and g are given below.
compute the derivatives. if an answer does not exist, type dne.
$\frac{d}{dx}f(g(x))|_{x = 2}=$
$\frac{d}{dx}f(g(x))|_{x = 9}=$
$\frac{d}{dx}g(f(x))|_{x = 9}=$

Explanation:

Step1: Recall the chain rule

The chain rule states that \(\frac{d}{dx}f(g(x)) = f^{\prime}(g(x))\cdot g^{\prime}(x)\) and \(\frac{d}{dx}g(f(x))=g^{\prime}(f(x))\cdot f^{\prime}(x)\)

Step2: Find \(g(2)\) and \(g^{\prime}(2)\)

From the graph of \(g(x)\), when \(x = 2\), \(g(2)=4\).
The slope of \(g(x)\) for \(x\in[0,6]\) (since \(x = 2\) is in this interval) is \(m=\frac{10 - 1}{6-0}=\frac{3}{2}\), so \(g^{\prime}(2)=\frac{3}{2}\)
The slope of \(f(x)\) for \(x\in[0,4]\) (since \(g(2) = 4\) is the right - hand endpoint of the first - part of \(f(x)\)’s graph). The slope of \(f(x)\) for \(x\in[0,4]\) is \(m=\frac{0 - 8}{4-0}=- 2\), so \(f^{\prime}(4)=-2\)
Then \(\frac{d}{dx}f(g(x))\big|_{x = 2}=f^{\prime}(g(2))\cdot g^{\prime}(2)=f^{\prime}(4)\cdot g^{\prime}(2)=(-2)\times\frac{3}{2}=-3\)

Step3: Find \(g(9)\) and \(g^{\prime}(9)\)

From the graph of \(g(x)\), when \(x = 9\), \(g(9)=6\)
The slope of \(g(x)\) for \(x\in[6,13]\) (since \(x = 9\) is in this interval) is \(m=\frac{2 - 10}{13 - 6}=-\frac{8}{7}\), so \(g^{\prime}(9)=-\frac{8}{7}\)
The slope of \(f(x)\) for \(x\in[4,10]\) (since \(g(9)=6\) is in this interval) is \(m=\frac{3 - 0}{10 - 4}=\frac{1}{2}\), so \(f^{\prime}(6)=\frac{1}{2}\)
Then \(\frac{d}{dx}f(g(x))\big|_{x = 9}=f^{\prime}(g(9))\cdot g^{\prime}(9)=f^{\prime}(6)\cdot g^{\prime}(9)=\frac{1}{2}\times(-\frac{8}{7})=-\frac{4}{7}\)

Step4: Find \(f(9)\) and \(f^{\prime}(9)\)

From the graph of \(f(x)\), when \(x = 9\), \(f(9)=3\)
The slope of \(f(x)\) for \(x\in[4,10]\) (since \(x = 9\) is in this interval) is \(m=\frac{3 - 0}{10 - 4}=\frac{1}{2}\), so \(f^{\prime}(9)=\frac{1}{2}\)
The slope of \(g(x)\) for \(x\in[0,6]\) (since \(f(9) = 3\) is in this interval) is \(m=\frac{10 - 1}{6-0}=\frac{3}{2}\), so \(g^{\prime}(3)=\frac{3}{2}\)
Then \(\frac{d}{dx}g(f(x))\big|_{x = 9}=g^{\prime}(f(9))\cdot f^{\prime}(9)=g^{\prime}(3)\cdot f^{\prime}(9)=\frac{3}{2}\times\frac{1}{2}=\frac{3}{4}\)

Answer:

\(\frac{d}{dx}f(g(x))\big|_{x = 2}=-3\)
\(\frac{d}{dx}f(g(x))\big|_{x = 9}=-\frac{4}{7}\)
\(\frac{d}{dx}g(f(x))\big|_{x = 9}=\frac{3}{4}\)