Find the slope of the tangent line to the graph of $f^{-1}$ Given function $f$, find the slope of the line tangent to the graph $f^{-1}$ at the point on the graph $f^{-1}$. $f(x)=\sqrt{5x}$; $(4,\frac{16}{5})$?
Here is what I have thus far:
$f'(x)= \frac{\sqrt{5}}{2\sqrt{x}}$ 
can it also be $f'(x)= \frac{2.5}{\sqrt{5x}}$?
We then need to plug in for $x$ at $x=\frac{16}{5}$
$f'(x)= \frac{\sqrt{5}}{2\sqrt{\frac{16}{5}}}=\frac{5}{8}$
We then have to reciprocate $\frac{5}{8}$
Let $M$ be the slope of the tangent line $f^{-1}$
$M=\frac{8}{5}$
Are these steps and arithmetic correct?
 A: This is wrong. Because $f'(a)=b$ doesn't mean at all that $(f^{-1})'(a)=\frac{1}b$
(quick counterexample on $\Bbb{R}^+_*$ for $x=1$ : take $f(x)=e^x,f^{-1}(x)=\ln x$,$f'(1)=e,(f^{-1})'(1)=1$)
What you should do : find the inverse function of $f$, then compute its derivative the given point.
Hint to find $f^{-1}$ : the inverse function of $x^2$ is $\sqrt{x}$
A: y= f(x) = $\sqrt{5x}$ $\rightarrow$ $y^2$ = $5x$ $\rightarrow$ $\frac{1}{5}y^2$ = x 
So $f^{-1}(x)$ = $\frac{1}{5}x^2$ , which is easily verified by taking $f^{-1}(f(x))$= x. 
Now: 
$(f^{-1})'(x)$ = $\frac{2}{5}x$. The slope of the tangent line of $f^{-1}$ at (4,$\frac{16}{5}$) is $(f^{-1})'(4)$ = $\frac{2}{5}*4$ =$\frac{8}{5}$. 
So the tangent line to the curve of $(f^{-1})'(x)$ at ( 4,$\frac{16}{5}$) is y-$\frac{16}{5}$ = $\frac{8}{5}$ (x - 4) $\rightarrow$ y = $\frac{8}{5}x$ -$\frac{16}{5}$. 
A graph of this situation is as follows: 

This is the point slope form of the tangent line. 
A: look at what the function $$y=f(x) = (5x)^{1/2} \text{ does at } x = 16/5, y = 4.$$ the derivative of $f$ is $$f'(x) = \frac12 (5x)^{-1/2}\times  5,\quad  
f'(16/5) = \frac 52 \times \frac 14= \frac58 $$
so far we have at $$x = {16}/{5}, y=f(16/5) = 4, f'(16/5) = \frac58.$$ therefore
$$ y = 4, f^{-1}(4) = 16/5, \left(f^{-1}\right)'(4) = \frac85$$
