# Evaluate: \frac{d}{dx} \int_0^{x^2} e^{-t^2}dt

## Question:

Evaluate: {eq}\frac{d}{dx} \int_0^{x^2} e^{-t^2}dt{/eq}

## Fundamental Theorem of Calculus

We can take the derivative of a variety of functions, even those defined as indefinite integrals. In fact, we can even apply the Fundamental Theorem of Calculus to take the derivative of an integral without actually working out any antiderivatives. Depending on the bounds of integration, we may need to also use the Chain Rule in this procedure.

{eq}\frac{d}{dx} \int_{g(x)}^{h(x)} f(t) dt = f(h(x)) h'(x) - f(g(x)) g'(x) {/eq}

The Fundamental Theorem of Calculus can be applied to simplify this problem greatly. That's because this theorem defines a shortcut for when we take the derivative of an integral. The integral defined in this problem has a function of x as its upper bound, so we'll need to also use the Chain Rule in this computation. However, since the lower bound is a constant, this will disappear during differentiation. Thus, we only need to concern ourselves with the upper bound.

The computation can thus be worked out as follows.

{eq}\begin{align*} \frac{d}{dx} \int_0^{x^2} e^{-t^2}dt &= e^{-(x^2)^2} \cdot \frac{d}{dx} x^2\\ &= 2xe^{-x^4} \end{align*} {/eq}