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Let $f$ be a one to one function defined on an interval, and suppose that $f$ is differentiable at $f^{-1}(b)$, with the derivative $f'(f^{-1}(b))\neq 0$. Then prove that $f^{-1}$ is differentiable at $b$ and $(f ^{-1})'(b)=\frac {1}{f'(f^{-1}(b))}$.

I guess, if $b=f(a)$ and $a=f^{-1}(b)$. but I couldn't get any idea to move further.

thanks

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    Start from the definition of differentiability. What limit must exist for $f^{-1}$ to be differentiable at $b$? – 2017-02-03
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    See this http://math.stackexchange.com/a/2114577/72031 – 2017-02-03
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    Without additional assumptions, $f^{-1}$ may even fail to be continuous at $b.$ – 2017-02-05

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Fix $\varepsilon>0$, small enough (concretely $\varepsilon0$ such that $$ \left|\frac{f(y)-f(a)}{y-a}-f'(a)\right|<\frac{\varepsilon\,f'(a)}{2}$$ if $|y-a|<\delta$. Now, writing $a_1=f^{-1}(b+h)$, with $h$ small enough so that $|a_1-a|<\delta$ (this comes from the continuity of $f^{-1}$ at $b$, which follows from the continuity of $f$ at $a$), \begin{align} \left|\frac{f^{-1}(b+h)-f^{-1}(b)}{h}-\frac1{f'(a)}\right|&=\left|\frac{a_1-a}{f(a_1)-f(a)}-\frac1{f'(a)}\right|\\ \ \\ &=\left|\left(f'(a)-\frac{f(a_1)-f(a)}{a_1-a}\right)\frac1{f'(a)}\,\frac{1}{\frac{f(a_1)-f(a)}{a_1-a}}\right|\\ \ \\ &=\left|\left(f'(a)-\frac{f(a_1)-f(a)}{a_1-a}\right)\frac1{f'(a)}\,\frac{1}{f'(a)-\varepsilon}\right|\\ \ \\ &\leq\frac{\varepsilon\,f'(a)}{2(f'(a)-\varepsilon)^2} \leq\frac{\varepsilon\,f'(a)}{2f'(a)/2}=\varepsilon. \end{align}

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    @@Martin, Is this complete. ? or should I have to do something further? – 2017-02-03
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    Well, what is glossed over in my argument is how you justify that if $h$ is small enough, then you can guarantee that $|a_1-a|<\delta$. Now, if you have to ask... – 2017-02-03
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    The hypotheses do NOT imply that $f^{-1}$ is continuous at $b.$ – 2017-02-03
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    @user254665: Please show me an example of a one-to-one function $f $, continuous at $a=f^{-1}(b) $, and such that $f^{-1} $ is not continuous at $b $. – 2017-02-03
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    Let $0=a=f(b)=b.$ Let $f(x)=x$ for $x<0.$..... For $ n\in \mathbb N$ and $x\in [2^{-n-1},2^{-n})$ let $f(x)=x-10^{-n}(x-2^{n+1}).$..... Let $f$ map $[1/2,1]$ bijectively onto $\cup_{n\in \mathbb N}[2^{-n}(1-10^{-n}/2),2^{-n}).$...... Now for x\in [2^{-n-1},2^{-n} we have $1\geq(f(x)-f(0))/(x-0)=f(x)/x=1-10^{-n}(x-2^{-n-1})/x> 1-10^{-n} 2^{-n-1}/x\geq 1-10^{-n}.$ Now $n\to \infty$ as $x\to 0.$ So $f'(0)=1.$............. As to the inverse function, every nbhd of $0$ has a point $y$ such that $f^{-1}y\in [1/2,1]$ but $f^{-1}(0)=0 $ so $f^{-1}$ is discontinuous at $0.$ – 2017-02-05