# I need help finding the derivative of the inverse function.

So $$f(x)=\frac{x+1}{2x-1}$$ and $$g(x)$$ is an inverse of $$f(x)$$

I have the points on $f(x)$ of (2,1). So I know that $f(2)=1$, $g(1)=2$ and $g'(1)=\frac{1}{f'[g(1)]}$ so $g'(1)=\frac{1}{f'(2)}$ and $f'(x)=\frac{(2x-1)-2(x+1)}{(2x-1)^2}$ so after plugging in 2 in $f'(x)$ I get $\frac{-3}{9} or -\frac{1}{3}$. However, I'm getting the wrong answer and I do not know what I am doing wrong. I have one more attempt at getting the correct answer, so any help would be greatly appreciated. A step by step explanation of what I'm doing wrong would also be greatly appreciated, thanks!

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user99680 made a nice observation, but actually you didn't do anything wrong except an incorrect manipulation. You found $f'(2) = -\frac{1}{3}$, and you want $g'(1) = \frac{1}{f'(2)}$ which would be $-3$.
Notice that the function is its own inverse, i.e., $f(f(x))=x$: $$\frac {f(x)+1}{2f(x)-1} = \frac {\frac{x+1}{2x-1}+1}{\frac{2x+2}{4x-2}-1}=\frac{\frac{x+1+2x-1}{2x-1}}{\frac{2x+2-4x+2}{4x-2}}=\frac{\frac{3x}{2x-1}}{\frac{2(3x)}{2(x-1)}}=x$$. Since $f$ is its own inverse, $(f^{-1}(1))'=f'(1)=-3$, since $f'(x)= \frac {-3}{(2x-1)^2}$, as you computed yourself.