# A student determines that \frac{d}{dx} \int ^{x^{2}}_{0} \cos^{2}(t)dt=\cos^{2}(x^{2}) by...

## Question:

A student determines that

{eq}\frac{d}{dx} \int ^{x^{2}}_{0} \cos^{2}(t)\;dt=\cos^{2}(x^{2}) {/eq} by applying the Fundamental Theorem of Calculus, is the student correct?

## Fundamental Theorem of Calculus; Chain Rule:

To differentiate the integral function {eq}\int_{0}^{x^2}\cos^{2}(t)\,dt, {/eq} we'll use the Fundamental Theorem of Calculus and the Chain Rule. For this we have the general statement:

For any continuous function {eq}f, {/eq} and any differentiable function {eq}g, {/eq} we have:

{eq}\begin{align*} \frac{d\left(\int_{0}^{g(x)}f(t)\,dt\right)}{dx}&=f(g(x))\frac{d(g(x))}{dx}\\ &=g'(x)\cdot f(g(x)). \end{align*} {/eq}

The student got the answer wrong, even though the Fundamental Theorem of Calculus is a correct tool to find the derivative. However, we need to use the chain rule to deal with the upper limit variable which is {eq}x^2 {/eq} rather than {eq}x {/eq} as in the regular Fundamental Theorem of Calculus.

In fact, to generalize the Fundamental Theorem of Calculus we can say the following:

For any continuous function {eq}f, {/eq} and any differentiable function {eq}g, {/eq} we have:

{eq}\begin{align*} \frac{d\left(\int_{0}^{g(x)}f(t)\,dt\right)}{dx}&=f(g(x))\frac{d(g(x))}{dx}\\ &=g'(x)\cdot f(g(x)). \end{align*} {/eq}

Thus, to find the correct derivative in this problem, we can find the derivative of the integral {eq}\int_{0}^{x^2}\cos^{2}(t)\,dt {/eq} using the Fundamental Theorem of Calculus and the Chain Rule:

{eq}\begin{align*} \frac{d\left(\int_{0}^{x^2}\cos^{2}(t)\,dt\right)}{dx}&=\cos^{2}(x^2)\cdot\frac{d(x^2)}{dx}\\ &=\boxed{2x\cos^{2}(x^2)}. \end{align*} {/eq}

Observation: The theorem mentioned above can be proved as follows:

We recall the Fundamental Theorem of Calculus telling us that if:

{eq}F(x)= \int_{a}^{x} f(t) \; dt, {/eq}

then {eq}F'(x)=f(x) {/eq}.

Combining the equation above with the Chain Rule for derivatives, we get that if:

{eq}h(x)=F(g(x))= \int_{a}^{g(x)} f(t) \; dt, {/eq}

then

{eq}g'(x)=\frac{d\left( F(g(x)) \right)}{dx}=f(g(x))\cdot g'(x) . {/eq}