# How do you use the first part of the Fundamental Theorem of Calculus to find the derivative of y = int 3(sin(t))^4 dt from e^x to 1?

Jul 23, 2018

The derivative is $- 3 {e}^{x} {\left(\sin \left({e}^{x}\right)\right)}^{4}$.

#### Explanation:

$y = {\int}_{{e}^{x}}^{1} 3 {\left(\sin \left(t\right)\right)}^{4} \mathrm{dt}$

Let's let $f \left(t\right) = 3 {\left(\sin \left(t\right)\right)}^{4}$.

Remember that by the fundamental theorem of calculus, a definite integral of a function is calculated using its antiderivative.

${\int}_{a}^{b} f \left(t\right) \mathrm{dt} = F \left(b\right) - F \left(a\right)$ where $F$ is an antiderivative of $f$.

Similarly:

${\int}_{{e}^{x}}^{1} f \left(t\right) \mathrm{dt} = F \left(1\right) - F \left({e}^{x}\right)$ where $F$ is an antiderivative of $f$.

$y = F \left(1\right) - F \left({e}^{x}\right)$

Because the question asks for it, we take the derivative.

$\textcolor{red}{\frac{d}{\mathrm{dx}}} y = \textcolor{red}{\frac{d}{\mathrm{dx}}} \left(F \left(1\right)\right) - \textcolor{red}{\frac{d}{\mathrm{dx}}} \left(F \left({e}^{x}\right)\right)$

Don't be scared. $F \left(1\right)$ is just a constant.

$\frac{\mathrm{dy}}{\mathrm{dx}} = 0 - F ' \left({e}^{x}\right) \cdot \frac{d}{\mathrm{dx}} \left({e}^{x}\right)$

But by definition of an antiderivative, $F ' \left(x\right) = f \left(x\right)$.

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {e}^{x} f \left({e}^{x}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - {e}^{x} \left(3 {\left(\sin \left({e}^{x}\right)\right)}^{4}\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - 3 {e}^{x} {\left(\sin \left({e}^{x}\right)\right)}^{4}$