Question

1. y = √f(x), where f is differentiable and nonnegative at x
2. y = ⁵√f(x)g(x), where f and g are differentiable at x

Explanation:

Step1: Recall chain - rule for 75

The chain - rule states that if $y = u^{n}$ and $u = f(x)$, then $\frac{dy}{dx}=n\cdot u^{n - 1}\cdot u'$. For $y=\sqrt{f(x)}=(f(x))^{\frac{1}{2}}$, let $u = f(x)$.
$\frac{dy}{dx}=\frac{1}{2}(f(x))^{-\frac{1}{2}}\cdot f'(x)=\frac{f'(x)}{2\sqrt{f(x)}}$

Step2: Recall product - rule and chain - rule for 76

First, rewrite $y=(f(x)g(x))^{\frac{1}{5}}$. Let $u = f(x)g(x)$. By the product - rule, $u'=f'(x)g(x)+f(x)g'(x)$. By the chain - rule, $\frac{dy}{dx}=\frac{1}{5}(u)^{-\frac{4}{5}}\cdot u'$.
Substitute $u = f(x)g(x)$ and $u'=f'(x)g(x)+f(x)g'(x)$ into the chain - rule formula:
$\frac{dy}{dx}=\frac{f'(x)g(x)+f(x)g'(x)}{5\sqrt[5]{(f(x)g(x))^{4}}}$

Answer:

For 75: $\frac{f'(x)}{2\sqrt{f(x)}}$
For 76: $\frac{f'(x)g(x)+f(x)g'(x)}{5\sqrt[5]{(f(x)g(x))^{4}}}$