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?
 A: $g(x)$ is the inverse of $f(x)$ means that $f \circ g(x)=f(g(x))=x=g(f(x))$
You take the derivative and get
$g'(x) \times f'(g(x))=1$ and also $f'(x) \times g'(f(x))=1$
A: 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.
A: Hint: First solve for $x_0$ such that: $2x^3 + 7x - 1 = \frac{191}{4}$, then apply inverse function theorem for derivative: $g'(f(x_0)) = \dfrac{1}{f'(x_0)}$
