Question

if (v(t)=sqrt{t^{7}}-sqrt{t}), then find the second derivative, (v(t)=)

Explanation:

Step1: Rewrite the function

Rewrite $v(t)=\sqrt{t^{7}}-\sqrt{t}$ as $v(t)=t^{\frac{7}{2}}-t^{\frac{1}{2}}$.

Step2: Find the first - derivative

Using the power - rule $\frac{d}{dt}(t^{n}) = nt^{n - 1}$, we have $v'(t)=\frac{7}{2}t^{\frac{7}{2}-1}-\frac{1}{2}t^{\frac{1}{2}-1}=\frac{7}{2}t^{\frac{5}{2}}-\frac{1}{2}t^{-\frac{1}{2}}$.

Step3: Find the second - derivative

Apply the power - rule again. $v''(t)=\frac{7}{2}\times\frac{5}{2}t^{\frac{5}{2}-1}-\frac{1}{2}\times(-\frac{1}{2})t^{-\frac{1}{2}-1}=\frac{35}{4}t^{\frac{3}{2}}+\frac{1}{4}t^{-\frac{3}{2}}$.

Answer:

$\frac{35}{4}t^{\frac{3}{2}}+\frac{1}{4}t^{-\frac{3}{2}}$