# Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function. F(x) =...

## Question:

Use Part 1 of the Fundamental Theorem of Calculus to find the derivative of the function.

{eq}F(x) = \int_{x}^{0} \, \sqrt{1 + \sec(8t)} \, \mathrm{d}t {/eq}

## Definite Integrals:

The fundamental formula that is used to find the derivative of the integral expression of the form: {eq}\int_{k(x)}^{a} f(t) dt {/eq} will be given as : {eq}\frac{d\left ( \int_{k(x)}^{a} f(t) dt \right )}{dx}=-\frac{d(k(x))}{dx} f(k(x)) \\ {/eq} The derivative of the integral is used with the theory of limits too, but for the derivative of the integral expression, we use the formula given above.

In this question we are having the integral expression with lower limits as a variable term.

So we have an integral expression:

{eq}F(x) = \int_{x}^{0} \, \sqrt{1 + \sec(8t)} \, \mathrm{d}t {/eq}

The derivative will be found as follows:

{eq}\frac{d(F(x))}{dx}= \frac{d\left (\int_{x}^{0} \, \sqrt{1 + \sec(8t)} \, \mathrm{d}t \right )}{dx}\\ {/eq}

There is fundamental theorem that is used here as per the formula:

{eq}\frac{d\left ( \int_{k(x)}^{a} f(t) dt \right )}{dx}=-\frac{d(k(x))}{dx} f(k(x)) \\ {/eq}

So the given expression will be now;

{eq}\frac{d(F(x))}{dx}= \frac{d\left (\int_{x}^{0} \, \sqrt{1 + \sec(8t)} \, \mathrm{d}t \right )}{dx}\\\\ =- \frac{d(x)}{dx} \sqrt{1 + \sec(8x)} \\ =-\sqrt{1 + \sec(8x)}\\ \Rightarrow F'(x)= -\sqrt{1 + \sec(8x)}\\ {/eq}

So this is the required derivative. 