Derivative of $d^2x/dy^2$

Question

Using the quotient rule and the chain rule you get $$ \frac{d}{dy}\left(\frac{dx}{dy}\right)=\frac{d}{dy}\left(\frac{1}{\frac{dy}{dx}}\right) =\frac{d}{dx}\left(\frac{1}{\frac{dy}{dx}}\right)\times \frac{dx}{dy}\\$$

I don't understand how this came. First is dy/dx always gonna be equal to dx/dy? Wouldn't the function have to be inverse to each other. And how to do chain rule in cases like this... I didn't get how the differentiation is done here. Can you explain with a simple example how chain differentiation like these are done. I am not too good at differentiation

Answer 1

I am confused about what you have shown and what you are interested to know...

However,the expression you have given,I think these are the alternative forms of $\dfrac{d^2 x}{dy^2}$.

Basically, we need to apply chain rule when there is a composite function. Say, $$y=f(x)=(x+1)^2$$ $$Let,g(x)=x+1$$ $$\implies y=f(x)=(g(x))^2$$ $$\implies \dfrac{dy}{dx}=2g(x).\bigg(\dfrac{d}{dx}g(x)\bigg)$$ $$\implies \dfrac{dy}{dx}=2(x+1).1=2(x+1)$$

What you have shown,I think some simplification (other forms) of double differentiation.So,you will differentiate two times respectively and will use chain rule if needed.

Answer 2

Take $y=x^2$, which is $x=\sqrt y$.

Then

$$\frac{dx}{dy}=\frac1{2\sqrt y}=\frac1{\dfrac{dy}{dx}}=\frac1{2x}$$ seems true.

Next,

$$\frac d{dy}\left(\frac{dx}{dy}\right)=-\frac1{4y\sqrt y}=\frac d{dy}\left(\frac1{\dfrac{dy}{dx}}\right)=\frac d{dy}\left(\dfrac1{2x}\right)=\frac d{dx}\left(\dfrac1{2x}\right)\frac{dx}{dy}=-\frac1{2x^2}\frac1{2x}=-\frac1{4x^3}$$

also looks right.