1. ## Differentiating the inverse

I took notes in class, but am totally stuck now. I need to find the inverse function of f(x)=4x+10x^9 at c= -14

2. Originally Posted by bnation
I took notes in class, but am totally stuck now. I need to find the inverse function of f(x)=4x+10x^9 at c= -14
no, you don't need the inverse of $\displaystyle f(x) = 4x + 10x^9$.

if $\displaystyle g(x)$ is the inverse of $\displaystyle f(x)$, then $\displaystyle f[g(x)] = x$

$\displaystyle \frac{d}{dx}(f[g(x)] = x)$

$\displaystyle f'[g(x)] \cdot g'(x) = 1$

$\displaystyle g'(x) = \frac{1}{f'[g(x)]}$

note that $\displaystyle f(-1) = -14$

since $\displaystyle g(x)$ is the inverse, $\displaystyle g(-14) = -1$

so ...

$\displaystyle g'(-14) = \frac{1}{f'[g(-14)]}$

finish up and find $\displaystyle g'(-14)$.

3. Originally Posted by bnation
I took notes in class, but am totally stuck now. I need to find the inverse function of f(x)=4x+10x^9 at c= -14
Alternatively, remember that for an inverse, the domain and ranges swap. I.e, the $\displaystyle x$ and $\displaystyle y$ values swap.

So if $\displaystyle f(x) = y = 4x + 10x^9$ then $\displaystyle f^{-1}(x) = x = 4y + 10y^9$.

You can differentiate the inverse using implicit differentiation.

$\displaystyle x = 4y + 10y^9$

$\displaystyle \frac{d}{dx}(x) = \frac{d}{dx}(4y + 10y^9)$

$\displaystyle 1 = \frac{dy}{dx}\,\frac{d}{dy}(4y + 10y^9)$

$\displaystyle 1 = \frac{dy}{dx}(4 + 90y^8)$

$\displaystyle \frac{dy}{dx} = \frac{1}{4 + 90y^8}$.

It'd be really tough to write the derivative in terms of $\displaystyle x$ though, so you don't bother.

If you're trying to evaluate this derivative at $\displaystyle x = -14$ then find what $\displaystyle y$ equals there, then substitute the $\displaystyle y$ value into the derivative.