Question

Use the Fundamental Theorem to find the derivative of this function.

The integral of arctan(6t) dt from 1/x to x

Answer 1

What is \(\dfrac{d}{dt}arctan(6t)\)?

Answer 2

\[\begin{align*}\frac{d}{dx}\left[\int_{1/x}^x\arctan6t~dt\right]&=\frac{d}{dx}\left[\int_{1/x}^c\arctan6t~dt\right]+\frac{d}{dx}\left[\int_c^x\arctan6t~dt\right]\\
&=-\frac{d}{dx}\left[\int_c^{1/x}\arctan6t~dt\right]+\frac{d}{dx}\left[\int_c^x\arctan6t~dt\right]
\end{align*}\]
Apply the FTC:
\[\frac{d}{dx}\left[\int_c^{g(x)}f(t)~dt\right]=f(g(x))\cdot g'(x)\]