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Suppose that $f$ is a one-to-one function and that $f^{−1}$ has a derivative which is nowhere $0$. Prove that $f$ is differentiable.

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As a side-note, the cubic root on $\mathbb R$ is a counterexample if the derivative of $f^{-1}$ is zero somewhere. Its inverse is $x^3$ which is differentiable everywhere, but the cubic root isn't. –  Najib Idrissi Jun 16 '13 at 7:54
    
Thanks alot @nik –  sammath Jun 16 '13 at 7:59
    
This question is phrased as an imperative sentence ("Prove that") rather than as a question. Many consider questions phrased in the imperative to be impolite. Moreover, you have left out two key things in the question: where did you encounter it (what book, what course?) and more importantly: what have you already tried? –  Carl Mummert Jun 16 '13 at 16:04
    
Thanks.I am bad at english. In my first course on Calculus.Can you pls frame it correctly for me? –  sammath Jun 16 '13 at 16:35

2 Answers 2

up vote 2 down vote accepted

Since $f$ is one-to-one, $f^{−1} $ is a function whose inverse is $f$.

Since $f^{−1} $ is differentiable, $f^{−1} $ is continuous.

Let $y = f(x)$ , so $f^{−1} (y) = x$.

$$\lim_{h\to0}\frac{f(x+h)−f(x)}{h}=\lim_{h\to0}\frac{f(x+h)−y}{h}.$$

Now since $f^{−1}$ is continuous, we can write $x + h = f^{−1} (y + k)$ where $k \to 0 $ as $ h \to 0.$

Now

$$\lim_ {h\to0} \frac {f(x+h)−y}{h} = \lim_{h\to0} \frac {f(f^{−1} (y+k))−y}{(x+h)−x} = \lim_{h\to0} \frac {(y+k)−y}{f^{−1} (y+k)−f^{−1} (y)} = \lim _{k\to0} \frac {k}{f^{−1} (y+k)−f^{−1} (y)} $$

as "$h \to 0 \implies k \to 0$".

Hence

$$ \lim_{h\to0} \frac {f(x+h)−y}{h} = \lim _ {k\to0}\frac {k} { f^{−1}(y+k) − f^{−1}(y) } = \frac {1} {(f^{−1})′(y) } = \frac {1} {(f^{−1})′(f(x)) }.$$

So f is differentiable.

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Another approach would be to use the formula for the derivative of $f^{-1}$, which is known to be:

$$\frac{df^{-1}}{dx}=\frac{1}{f^\prime(f^{-1}(x))}$$

We know that $\left(f^{-1}(x)\right)^{-1}=f(x)$

Ergo:

$$\frac{df}{dx}=\frac{1}{\left(f^{-1}\right)^\prime(f(x))}$$

And since the derivative of $f^{-1}$ is never $0$, then the function is differentiable everywhere.

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