## Calculus: Early Transcendentals 8th Edition

$$h'(1)=\frac{6}{5}$$
$$h(x)=\sqrt{4+3f(x)}$$ Apply the Chain Rule: $$h'(x)=\frac{d(\sqrt{[4+3f(x)]})}{d[4+3f(x)]}\frac{d[4+3f(x)]}{df(x)}\frac{df(x)}{dx}$$ $$h'(x)=\frac{1}{2\sqrt{4+3f(x)}}\times(0+3\times1)\times f'(x)$$ $$h'(x)=\frac{3f'(x)}{2\sqrt{4+3f(x)}}$$ Then, $$h'(1)=\frac{3f'(1)}{2\sqrt{4+3f(1)}}$$ We know $f(1)=7$ and $f'(1)=4$. Therefore, $$h'(1)=\frac{3\times4}{2\sqrt{4+3\times7}}=\frac{12}{2\sqrt{25}}=\frac{6}{5}$$