Question

A table of values for f, g, f', and g' is given
x f(x) g(x) f'(x) g'(x)
1 2 4 3 -7
2 4 1 -5 8
4 1 2 6 -9
If h(x) = g(f(x^2)), find h'(2)

The answer is -168. I've tried everything. No matter how far I go in using chain rule, I keep getting different answers and none of them are -168. Help!

Answer 1

Answer:

As you've mentioned, $h'(2) = -168$. See explanation.

Step-by-step explanation:

$h(x) = g(f(x^{2}))$ is a composite function about $x$.

$\displaystyle \begin{aligned}h'(x) &= \frac{d}{dx}[g(f(x^{2}))] \\ &=g'(f(x^{2}) \cdot \frac{d}{dx}[f(x^{2})]&&\text{Chain rule; treat} \; f(x^{2})\;\text{as the inner function.} \\&=g'(f(x)^{2})\cdot f'(x^{2})\cdot \frac{d}{dx}[x^{2}] &&\text{Chain rule; treat} \; x^{2}\;\text{as the inner function.} \\ &= g'(f(x^{2}))\cdot f'(x^{2}) \cdot 2\;x&&\text{Power Rule.}\; \frac{d}{dx}[x^{2}] = 2\;x. \\&= 2\;x \cdot f'(x^{2})\cdot g'(f(x^{2}))\end{aligned}$.

For $x = 2$:

• $2\;x = 4$;
• $x^{2} = 4$;
• $f'(x^{2}) = f(4) = 6$;
• $f(x^{2}) = 1$;
• $g'(f(x^{2}) = g'(1) =-7$.

$\displaystyle h'(x) = 2\;x \cdot f'(x^{2})\cdot g'(f(x^{2})) = -168$.