# Find the derivative of the function. g(x) = integral from (5x) to (7x) (u^2 - 4)/(u^2 + 4) du.

## Question:

Find the derivative of the function.

{eq}g(x) = \int_{5x}^{7x} \, \frac{u^2 - 4}{u^2 + 4} \, \mathrm{d}u {/eq}

## Fundamental Theorem of Calculus

The Fundamental Theorem of Calculus (FTC) Part I is used to find the derivative of an antiderivative,

{eq}\displaystyle \frac{d}{dx}\left(\int_a^x f(t)dt\right) =f(x), \text{ where } a - \text{ is a constant},\\ \displaystyle \frac{d}{dx}\left(\int_a^{u(x)} f(t)dt\right) =u'(x)f(u(x)), \text{ (Chain Rule}). {/eq}

To apply FTC Part I, we need to make sure the lower limit of integration is a constant.

Sometimes, we need to use the following integration properties, when we apply FTC Part I:

{eq}\displaystyle \int_a^b f(t)dt=-\int_b^a f(t)dt\\ \displaystyle \int_a^b f(t)dt=\int_a^c f(t)dt+\int_c^b f(t)dt. {/eq}

To find the derivative of the function {eq}\displaystyle g(x) = \int_{5x}^{7x} \, \frac{u^2 - 4}{u^2 + 4} \, du, {/eq} we will apply the Fundamental Theorem of Calculus Part I, as below

{eq}\displaystyle \begin{align} g'(x)= \frac{d}{dx}\left[\int_{5x}^{7x} \, \frac{u^2 - 4}{u^2 + 4} \ du\right]&= \frac{d}{dx}\left[\int_{5x}^{0} \, \frac{u^2 - 4}{u^2 + 4} \ du+\int_{0}^{7x} \, \frac{u^2 - 4}{u^2 + 4} \ du\right], &\left[\text{ by splitting the integral at a constant }\right]\\ &= \frac{d}{dx}\left[-\int_{0}^{5x} \, \frac{u^2 - 4}{u^2 + 4} \ du+\int_{0}^{7x} \, \frac{u^2 - 4}{u^2 + 4} \ du\right], &\left[\text{ to bring the constant at the lower limit of integration}\right]\\ &= -\frac{d}{dx}\left[\int_{0}^{5x} \, \frac{u^2 - 4}{u^2 + 4} \ du\right]+\frac{d}{dx}\left[\int_{0}^{7x} \, \frac{u^2 - 4}{u^2 + 4} \ du\right]\\ &= -\left(\frac{(5x)^2 - 4}{(5x)^2 + 4} \right)\cdot 5+\left(\frac{(7x)^2 - 4}{(7x)^2 + 4} \right)\cdot 7, &\left[\text{by apply FTC Part I}\right]\\ &= \boxed{ -5\left(\frac{25x^2 - 4}{25x^2 + 4} \right)+7\left(\frac{49x^2 - 4}{49x^2 + 4} \right)}. \end{align} {/eq}