<s>[INST] 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?
 [/INST] 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
$$ y = 4, f^{-1}(4) = 16/5, \left(f^{-1}\right)'(4) = \frac85$$
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