Derivative of an inverse Let $f(x)=2x^3+7x−1$, and let $g(x)$ be the inverse of $f(x)$. Then find $g′(191/4)$. I know only one way of doing this. Solving the cubic equation for x and then differentiating it. This is too too long. How to solve it more easily?

In order to not confuse the variables of the inverse function with the variables of the function itself, let :
$y=f(x)=2x^3+7x−1=X$
$Y=g(X)=x$
One little picture says more than a long speech! : See below

$\frac{dg}{dX}=\frac{1}{\frac{df}{dx}}$
For $X=\frac{191}{4}$ we have $y=\frac{191}{4}=2x^3+7x-1$ which real root is $x=\frac{5}{2}$
$\frac{1}{\frac{df}{dx}}=\frac{1}{6x^2+7}=\frac{1}{6\left(\frac{5}{2}\right)^2+7}=\frac{2}{89}$ 
$\frac{dY}{dX}=\frac{1}{\frac{df}{dx}}=\frac{2}{89}$ 
where $X=y=\frac{191}{4}$  and $Y=x=\frac{5}{2}$
One see on the graph on right, which represents the inverse function, that the point considered is at abscissa $\frac{191}{4}$ as specified in the wording of the question.
