Derivative of inverse?

Question

Let $f$ be the function described implicitly by $$y \cos x = e^x \ln y + e - 1.$$

What is the derivative of $f^{-1}$ at $e$?

How to tackle this one? I don't need the result, please tell me a hint or method on how to find $f^{-1}$. Thanks

Answer 1

Using chain rule we see that:

$$\frac{d}{d (f(x))}f^{-1}(f(x)) = \frac{1}{f'(x)} \tag{1}$$

Here we see that for derivative $f^{-1}(x)$ at $x = e$, you need to find $x$ for which $f(x) =e$.

You ask yourself, for what value of $x$ does $f(x) = e$ ? From the given equation in $f(x)$ (that is, $y$), we let $y = e$ and solve for $x$:

$$e \cos x = e^x +e - 1$$

From here you see that $y = e$ when $x = 0$.

Now we have $f(x) = e$ at $x = 0$. Then using $(1)$ we have :

$$\frac{d }{dx} f^{-1}(x) |_{x =e} = \frac{1}{f'(0)} $$

Answer 2

Let $f:A\to B$, where $A,B \subseteq \mathbb{R}$ be a bijective function and let $f^{-1}:B \to A$ be its inverse. Remember that $$x=f(y) \iff f^{-1}(x)=y.$$ Notice that $$1=\frac{dx}{dx}=\frac{df^{-1}(y)}{dx}=\frac{f^{-1}(y)}{dy}\cdot \frac{dy}{dx}=f’(y)\cdot f’(x),$$ thanks to the Chain Rule, and hence $$(f^{-1})’(x)=\frac{1}{f’(y)}=\frac{1}{f’(f^{-1}(x))}.$$ In your problem, you need to find $$(f^{-1})(e)=\frac{1}{f’(f^{-1}(e))}.$$

Solving for $y=e$ yields $e\cos x=e^x+e-1$, which is true for $x=0$. Hence $f^{-1}(e)=0$. Your problem is then to find $1/f’(0)$.