Question

differentiate the function. v(t) = t^{-3/5}+t^{9} v(t) =

Explanation:

Step1: Recall power - rule

The power - rule for differentiation is $\frac{d}{dt}(t^n)=nt^{n - 1}$, where $n$ is a real number.

Step2: Differentiate each term

For the first term $t^{-\frac{3}{5}}$, using the power - rule, we have $\frac{d}{dt}(t^{-\frac{3}{5}})=-\frac{3}{5}t^{-\frac{3}{5}-1}=-\frac{3}{5}t^{-\frac{8}{5}}$. For the second term $t^9$, using the power - rule, we have $\frac{d}{dt}(t^9)=9t^{9 - 1}=9t^8$.

Step3: Combine the results

$V'(t)=\frac{d}{dt}(t^{-\frac{3}{5}})+\frac{d}{dt}(t^9)=-\frac{3}{5}t^{-\frac{8}{5}}+9t^8$.

Answer:

$-\frac{3}{5}t^{-\frac{8}{5}}+9t^8$