# How do you use the second fundamental theorem of Calculus to find the derivative of given int tan(t^4)-1) dt from [1, x^3]?

May 3, 2016

#### Answer:

Please see the explanation section below.

#### Explanation:

The second fundamental theorem of calculus tells us that

$g \left(u\right) = {\int}_{1}^{u} \left(\tan \left({t}^{4}\right) - 1\right) \mathrm{dt} = F \left(u\right) - F \left(1\right)$

where $F$ is an antiderivative of $\left(\tan \left({t}^{4}\right) - 1\right)$.

That is, $F ' \left(u\right) = \tan \left({u}^{4}\right) - 1$.

Nore that, since $f(1) is simply some constant, we have $g ' \left(u\right) = F ' \left(u\right) = \tan \left({u}^{4}\right) - 1$. We have been asked for the derivative of $g \left({x}^{3}\right)$, so we'll use the chain rule. $\frac{d}{\mathrm{dx}} \left(g \left({x}^{3}\right)\right) = g ' \left({x}^{3}\right) \cdot \frac{d}{\mathrm{dx}} \left({x}^{3}\right) = \left(\tan \left({\left({x}^{3}\right)}^{4}\right) - 1\right) \frac{d}{\mathrm{dx}} \left({x}^{3}\right)$$= 3 {x}^{2} \left(\tan \left({x}^{12}\right) - 1\right)\$