Slopes of inverse functions I have a question that states if $f(x) = x^3+3x-1$ from $(-\infty,\infty)$ calculate $g'(3)$using the formula $$ g'(x)= \left(\frac1{f'(g(x))}\right )$$
If I am thinking about this correctly does that mean that the slope of $f(x)$ is the reciprocal of its inverse function $g(x)$
If so then can I just differentiate $f(x)$, plug in 3, and find the reciprocal of that to find $g'(3)$?
$$f'(x) = 3x^2 + 3$$
$$f'(3) = 30$$
$$g'(3) = \frac1{30}$$
 A: We need to find $g(3)$. $f(g(3))=3$, so $g(3)$ is a root of $a^{3}+3a-1=3$ so $a^{3}+3a-4=0$ so $(a-1)(a^{2}+a+4)=0$.By the quadratic formula, $a^{2}+a+4$ has no real root (discriminant is $-3<0$) so $g(3)=1$. Now $f'(x)=3x^{2}+3$ so $f'(g(3))=f'(1)=6$. So $g'(3)=\frac{1}{6}$. I hope I am not mistaken! Edited : I am assuming $g$ and $f$ are inverse functions, here $f$ has an inverse that is also differentiable. Now, what you said is not comletely right, $g'(x)$ is the reciprocal of $f'$ evaluated at $g(x)$!
A: Not quite. First you gave the correct formula 
$$
g'(x) = \frac{1}{f'(g(x))} \quad (*)
$$
which by the way is a consequence of $f(g(x)) = x$ and taking the derivative of both sides.
You then applied the wrong formula
$$
g'(3) = \frac{1}{f'(3)}
\mbox{vs.} \frac{1}{f'(g(3))}
$$
which will not work in general except if $g(3) = 3$ which is not the case here:
$$
f(x) = x^3 + 3x - 1 = 3
$$
has the only real solution $x = 1$. So we have $f(1) = 3$ which means $g(3) = 1$.
This leads to 
$$
g'(3) = \frac{1}{f'(1)} = \frac{1}{3\cdot 1^2+3} = \frac{1}{6}
$$
Compare this with the graphs:

(Larger version of the image)
A: No.
Suppose $f(10)=70$, so $g(70)=10$.
And suppose $f'(10)=3$.
It follows that $g'(70)=\dfrac 1 3$.
It does not follow that $g'(10)=\dfrac 1 3$.
You do just take reciprocals, but you need to evaluate each function at the right point.
