find h′(t) if h(t)=8/t^1/2 - 8/t^4/7. h′(t)=

Explanation:

Step1: Rewrite the function

Rewrite $h(t)$ as $h(t)=8t^{-\frac{1}{2}}-8t^{-\frac{4}{7}}$ using the rule $\frac{1}{x^n}=x^{-n}$.

Step2: Apply power - rule for differentiation

The power - rule states that if $y = ax^n$, then $y^\prime=anx^{n - 1}$. For the first term, when $a = 8$ and $n=-\frac{1}{2}$, the derivative of $8t^{-\frac{1}{2}}$ is $8\times(-\frac{1}{2})t^{-\frac{1}{2}-1}=-4t^{-\frac{3}{2}}$. For the second term, when $a = 8$ and $n = -\frac{4}{7}$, the derivative of $8t^{-\frac{4}{7}}$ is $8\times(-\frac{4}{7})t^{-\frac{4}{7}-1}=-\frac{32}{7}t^{-\frac{11}{7}}$.

Answer:

$-4t^{-\frac{3}{2}}-\frac{32}{7}t^{-\frac{11}{7}}$