Question

differentiate $v = t^{2}+\frac{-6}{sqrt4{t^{3}}}$. $v=$

Explanation:

Step1: Rewrite the function

Rewrite $\frac{-6}{\sqrt[4]{t^{3}}}$ as $- 6t^{-\frac{3}{4}}$, so $v=t^{2}-6t^{-\frac{3}{4}}$.

Step2: Apply power - rule for differentiation

The power - rule states that if $y = ax^{n}$, then $y^\prime=anx^{n - 1}$. For $y = t^{2}$, $y^\prime=2t^{2 - 1}=2t$. For $y=-6t^{-\frac{3}{4}}$, $y^\prime=-6\times(-\frac{3}{4})t^{-\frac{3}{4}-1}=\frac{18}{4}t^{-\frac{7}{4}}=\frac{9}{2}t^{-\frac{7}{4}}$.

Step3: Combine the derivatives

$v^\prime$ is the sum of the derivatives of the two terms. So $v^\prime = 2t+\frac{9}{2}t^{-\frac{7}{4}}$.

Answer:

$2t+\frac{9}{2t^{\frac{7}{4}}}$