Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have been told that if $y=g(x)$ then $$\dfrac{d^2y}{dx^2}=\dfrac{dy}{dx}\cdot \left( \dfrac{d}{dy}\dfrac{dy}{dx} \right) $$ if this is true please can some one tell me how we get this result?

Here is my explination, please could you tell me if it is correct: let $$\dfrac{dy}{dx}=f(y)$$ where $y=g(x)$ therfore $$\dfrac{dy}{dx}=f(g(t))$$ $$\dfrac{d^2y}{dx^2}=g'(t)f'(g(t))$$ which implys $$\dfrac{d^2y}{dx^2}=\dfrac{dy}{dx}\cdot \left( \dfrac{d}{dy}\dfrac{dy}{dx} \right)$$

Is this correct?

share|cite|improve this question
Ok, now write down your identity for the function $y=\sin x$, so that we can understand it. – Siminore Apr 26 '14 at 12:04
up vote 1 down vote accepted

The simpler relation connecting the first derivatives of $g$ and its inverse is


This notation is sometimes misleading, as long as you don't exactly know what you mean, so I like to write it more explicitly as


Now with this knowledge you can derive such a relation for the second derivatives:


Now with the chain rule of differentiation we can write this as


Using $(1)$ we have $x'(\eta)=1/y'(x(\eta))$ and obtain

$$=y'(x)\left(\frac{d}{d\eta} y'(x(\eta))\right)_{\eta=y(x)}$$

If you change back to implicit notation, which forgets the arguments, this is


The key rule for such calculations is, that you have to express everything as a function of one variable, which in my case is $x$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.