Let f(x) = x - 5. If g(x) = [f(4h(x) + 3)]^2 | vedclass

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(B) Given $f(x) = x - 5$, so $f'(x) = 1$.$g(x) = [f(4h(x) + 3)]^2$.Using the chain rule, $g'(x) = 2[f(4h(x) + 3)] \cdot f'(4h(x) + 3) \cdot \frac{d}{d

Vedclass · Accepted Answer

$-224$

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(B) Given $f(x) = x - 5$, so $f'(x) = 1$.$g(x) = [f(4h(x) + 3)]^2$.Using the chain rule, $g'(x) = 2[f(4h(x) + 3)] \cdot f'(4h(x) + 3) \cdot \frac{d}{dx}(4h(x) + 3)$.$g'(x) = 2[f(4h(x) + 3)] \cdot 1 \cdot 4h'(x) = 8h'(x) \cdot f(4h(x) + 3)$.At $x = 1$, $g'(1) = 8h'(1) \cdot f(4h(1) + 3)$.Substitute $h(1) = 4$ and $h'(1) = -2$:$g'(1) = 8(-2) \cdot f(4(4) + 3) = -16 \cdot f(19)$.Since $f(x) = x - 5$, $f(19) = 19 - 5 = 14$.$g'(1) = -16 \cdot 14 = -224$.

Let $f(x) = x - 5$. If $g(x) = [f(4h(x) + 3)]^2$ and $h(1) = 4, h'(1) = -2$, then $g'(1) = .....$

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