# Calculate the derivative: \frac{d}{dx} \int_0^{5x^2} \frac{tdt}{t+40}

## Question:

Calculate the derivative: {eq}\frac{d}{dx} \int_0^{5x^2} \frac{tdt}{t+40}{/eq}

## Fundamental Theorem of Calculus

Let {eq}F(x) {/eq} be a differentiable function with derivative {eq}F'(x) = f(x). {/eq} Then the definite integral {eq}\displaystyle \int_a^b f(x) dx {/eq} can be evaluated by the Fundamental Theorem of Calculus and is equal to {eq}F(b) - F(a). {/eq}

Consider the function {eq}G(x) = \dfrac{d}{dx} \displaystyle \int_0^{5x^2} \dfrac{tdt}{t+40}. {/eq} Let {eq}F(x) {/eq} be an antiderivative of {eq}f(x) = \dfrac{x}{x+40}. {/eq} Then by the Fundamental Theorem of Calculus we have:

{eq}\displaystyle \int_0^{5x^2} \dfrac{tdt}{t+40} = F(5x^2) - F(0). {/eq}

Therefore, differentiating and using the chain rule yields:

{eq}G(x) = \dfrac{d}{dx} \displaystyle \int_0^{5x^2} \dfrac{tdt}{t+40} = F'(5x^2)(10 x) = 10 x f(5x^2) {/eq}

Substituting yields the desired result:

{eq}G(x) = \dfrac{5x^2}{5x^2+40} 10 x = \dfrac{10 x^3}{x^2 + 8}. {/eq}