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I am trying to find the derivative of $f \left( g^{-1} (x) \right)$. I know that it is given by $$f \left( g^{-1} (x) \right)'=\frac{f' \left( g^{-1} (x) \right)}{ g' \left( g^{-1} (x) \right)}$$ using chain rule.

But I am not able to prove it using first principle.

How can I do that using the first principle ?

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  • $\begingroup$ Please provide some context. What is $g^{-1}(x)$ for example? $\endgroup$ – Yuriy S Jan 21 '18 at 20:38
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    $\begingroup$ $g^{-1}$ is the functional inverse of $g$, I suppose. $\endgroup$ – Botond Jan 21 '18 at 20:39
  • $\begingroup$ what do you mean by first principle? $\endgroup$ – Guy Fsone Jan 26 '18 at 20:27
  • $\begingroup$ @GuyFsone Here is the link explaining first principle. I hope it helps brilliant.org/wiki/derivative-by-first-principle $\endgroup$ – tarun14110 Jan 26 '18 at 20:29
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Enforcing the change of variables $\color{black}{y=g^{-1}(x)~~~and ~~~~b=g^{-1}(a)} $ then $ y\to b$ as $x\to a$ and hence by firs principle we have

$$\begin{align} \left(f(g^{-1})\right)'(a) &=\lim_{x\to a}\frac{f(g^{-1})(x)-f(g^{-1})(a)}{x-a}\\&= \lim_{x\to a}\frac{f(g^{-1})(x)-f(g^{-1})(a)}{\color{blue}{g^{-1}(x)-g^{-1}(a)}}\cdot\frac{\color{blue}{g^{-1}(x)-g^{-1}(a)}}{x-a}\\&=\lim_{x\to a}\frac{f(g^{-1})(x)-f(g^{-1})(a)}{\color{blue}{g^{-1}(x)-g^{-1}(a)}}\cdot\frac{\color{blue}{g^{-1}(x)-g^{-1}(a)}}{\color{red}{g(g^{-1})(x)-g(g^{-1})(a)}} \\&=\lim_{y\to b}\frac{f(y)-f(b)}{\color{blue}{y-b}}\cdot\frac{\color{blue}{y-b}}{\color{red}{g(y)-g(b)}} = \color{black}{f'(b)\cdot \frac{1}{g'(b)}}\\&= \color{brown}{ \frac{f'(g^{-1}(a))}{g'(g^{-1}(a))}}\end{align}$$

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Suppose that $g$ is an invertible function. Then $g^{-1}\circ g = g\circ g^{-1}=id$.

Notice that $g(g^{-1}(x))=x$ and differentiate both sides using the chain rule on the left: $$g'(g^{-1}(x)).(g^{-1}(x))'=1 $$ Thus $$(g^{-1}(x))'=\frac{1}{g'(g^{-1}(x))} $$

Now differentiate $f(g^{-1}(x))$ using the chain rule again to get $$(f(g^{-1}(x)))'=f'(g^{-1}(x))(g^{-1}(x))' $$

But we've calculated the second term above so $$ (f(g^{-1}(x)))'=\frac{f'(g^{-1}(x))}{g'(g^{-1}(x))} $$

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  • $\begingroup$ thanks Ravi for the answer. You explained it very well. But I would prefer the proof using first principle. Actually I am trying to get better at first principle. Can you help me with this ? $\endgroup$ – tarun14110 Jan 21 '18 at 20:50
  • $\begingroup$ What do you mean by first principles? Assuming only the chain rule and the fact that $g$ and $g^{-1}$ are inverses, we proved this relation. Do you want a proof of the chain rule too? Then you should make that a new question. $\endgroup$ – RKD Jan 21 '18 at 22:14
  • $\begingroup$ Here is the link explaining that. brilliant.org/wiki/derivative-by-first-principle $\endgroup$ – tarun14110 Jan 26 '18 at 20:23

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