Question

1. if f(x)=x² - 4 and g is a differentiable function of x, what is the derivative of f(g(x))? (a) 2g(x) (b) 2g(x) (c) 2xg(x) (d) 2g(x)g(x) (e) 2g(x) - 4 20. the table above gives values of f,f,g, and g for selected values of x. if h(x)=f(g(x)), what is the value of h(1)? (a) -19 (b) -14 (c) 7 (d) 9

Explanation:

Step1: Recall the chain - rule

The chain - rule states that if $y = f(g(x))$, then $y'=f'(g(x))\cdot g'(x)$. Given $f(x)=x^{2}-4$, then $f'(x) = 2x$. So the derivative of $f(g(x))$ is $f'(g(x))\cdot g'(x)=2g(x)\cdot g'(x)$.

Step2: For the second question

We know that if $h(x)=f(g(x))$, by the chain - rule $h'(x)=f'(g(x))\cdot g'(x)$. We want to find $h'(1)$. First, find $g(1)$ from the table. When $x = 1$, $g(1)=3$. Then find $f'(g(1))$ and $g'(1)$. Since $g(1) = 3$, $f'(g(1))=f'(3)$. From the table, when $x = 3$, $f'(3)=7$. And when $x = 1$, $g'(1)=-2$. Then $h'(1)=f'(g(1))\cdot g'(1)=7\times(-2)=-14$.

Answer:

1. D. $2g(x)g'(x)$
2. B. - 14