## Calculus 8th Edition

$\frac{2}{\pi}$
We have a formula $f(x)$, and are looking for the derivative of its inverse formula, $h(x)$. The formula for the derivative of $h(x)$ evaluated at a specific point, (i.e. where a=3) is as follows $h'(a)=\frac{1}{f'(h(a))}$ The first step is to find $f'(x)$ respectively. By simple differentiation and the chain rule we get that $f'(x)=2x+\frac{\pi sec(\frac{\pi x}{2})tan(\frac{\pi x}{2})}{2}$ Secondly, we will use the a-value to find the x-value in $f(x)$ that corresponds to it. $3=3+x^{2}+tan(\frac{\pi x}{2})$ By observing, we notice that the only way that 3 could equal the other side is if the x-terms equal 0, and since x is multiplying in both of them, the x-value that satisfies the equation is 0. $3=3+0^{2}+tan(\frac{\ 0\times \pi}{2})$ $3=3+0+tan(0)$ $3=3+0+0$ $3=3$ Thus, we can now state that $h(3)=0$, since $f(0)=3$ $h'(3)=\frac{1}{f'(h(3))}$ $h'(3)=\frac{1}{f'(0)}$ To continue working with the formula, we must solve for $f'(0)$ $f'(0)=2(0)+\frac{\pi sec(\frac{\pi (0)}{2})tan(\frac{\pi (0)}{2})}{2}$ $f'(0)=0+\frac{\pi sec(0)tan(0)}{2}$ $f'(0)=0+\frac{\pi (1)}{2}$ $f'(0)=\frac{\pi}{2}$ If plugged into the equation, $h'(3)=\frac{1}{f'(0)}$ $h'(3)=\frac{1}{\frac{\pi}{2}}$ Thus resulting as $h'(3)=\frac{2}{\pi}$