Question

use the figures below to evaluate the indicated derivative, or state that it does not exist. if the derivative does not exist, enter dne in the answer blank. the graph to the left (in black) gives $f(x)$, while the graph to the right gives $g(x)$ (which is constant for values of $x$ greater than 120).
$\frac{d}{dx}f(g(x))|_{x = 100}=$ (if the derivative does not exist, enter dne.)

Explanation:

Step1: Recall the chain - rule

The chain - rule states that if $y = f(g(x))$, then $\frac{dy}{dx}=f^{\prime}(g(x))\cdot g^{\prime}(x)$. We need to find $g(100)$ and $g^{\prime}(100)$ first, and then $f^{\prime}(g(100))$.

Step2: Analyze the graph of $g(x)$

From the graph of $g(x)$, we can see that $g(x)$ is a linear function for $x\leq120$. The slope of the line for $g(x)$ can be calculated using the formula $m=\frac{y_2 - y_1}{x_2 - x_1}$. Let's take two points on the line of $g(x)$, say $(0,40)$ and $(120,160)$. Then $g^{\prime}(x)=\frac{160 - 40}{120-0}=1$ for $x < 120$. And $g(100)$: Using the point - slope form $y - y_1=m(x - x_1)$ with $m = 1$, $x_1 = 0$, $y_1 = 40$, when $x = 100$, $g(100)=40+1\times100 = 140$.

Step3: Analyze the graph of $f(x)$

We need to find $f^{\prime}(g(100))=f^{\prime}(140)$. For $x>120$, the function $f(x)$ is linear. The slope of the right - hand part of $f(x)$ (for $x > 120$) can be calculated. Let's take two points on the right - hand part of $f(x)$, say $(120,0)$ and $(160,160)$. The slope $f^{\prime}(x)=\frac{160-0}{160 - 120}=4$ for $x>120$.

Step4: Apply the chain - rule

Now, using the chain - rule $\frac{d}{dx}f(g(x))|_{x = 100}=f^{\prime}(g(100))\cdot g^{\prime}(100)$. Since $g^{\prime}(100)=1$ and $f^{\prime}(g(100))=f^{\prime}(140)=4$, then $\frac{d}{dx}f(g(x))|_{x = 100}=4\times1=4$.

Answer:

4