# Second derivative identity proof?

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?

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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

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

$$\frac{dy}{dx}\frac{dx}{dy}=1$$

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

$$y'(x)x'(y(x))=1.$$

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

$$y''(x)=\frac{d}{dx}y'(x)=\frac{d}{dx}\frac{1}{x'(y(x))}\qquad(1)$$

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

$$=\left(\frac{d}{d\eta}\frac{1}{x'(\eta)}\right)_{\eta=y(x)}y'(x)$$

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

$$=\frac{dy}{dx}\frac{d}{dy}\frac{dy}{dx}.$$

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

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