Question

determine $\frac{dy}{dx}$ if $y = int_{0}^{x}\frac{3e^{t}}{sqrt{2t^{3}+8}}dt$. attempt 1: 2 attempts remaining. submit answer

Explanation:

Step1: Apply fundamental theorem of calculus

If $y = \int_{a}^{x}f(t)dt$, then $\frac{dy}{dx}=f(x)$. Here $a = 0$ and $f(t)=\frac{3e^{t}}{\sqrt{2t^{3}+8}}$.

Step2: Find the derivative

By the fundamental theorem of calculus, $\frac{dy}{dx}=\frac{3e^{x}}{\sqrt{2x^{3}+8}}$.

Answer:

$\frac{3e^{x}}{\sqrt{2x^{3}+8}}$