4

The actual question is :
$$\frac{d^2x}{dy^2}$$ equals :

$$1. \frac{d^2y}{dx^2}^{-1}$$

$$2. -\left(\frac{d^2y}{dx^2}\right)^{-1}\left(\frac{dy}{dx}\right)^{-3}$$

$$3.\left(\frac{d^2y}{dx^2}\right)\left(\frac{dy}{dx}\right)^{-2}$$

$$4.-\left(\frac{d^2y}{dx^2}\right)\left(\frac{dy}{dx}\right)^{-3}$$

I know that the answer is 4 , in fact the solution given is as follows : $d^2x/dy^2=-(d^2y/dx^2)(dy/dx)^{-2}(dx/dy)=$ option 4

I don't understand this.From where did the minus sign come? Also I don't understand how the above steps are true?
This is the first time I'm solving a problem like this . Please help.

Kamal Saleh
  • 6,497
A Googler
  • 3,355

2 Answers2

8

Recalling that $\dfrac{dx}{dy} = \dfrac{1}{\dfrac{dy}{dx}}$, we see that

$$\frac{d^2x}{dy^2} = \frac{d}{dy}\left(\frac{dx}{dy}\right) = \frac{d}{dy}\left(\frac{1}{\dfrac{dy}{dx}}\right)$$

Now, by quotient rule, we have that

$$\begin{aligned}\frac{d}{dy}\left(\frac{1}{\dfrac{dy}{dx}}\right) &= \frac{-\dfrac{d}{dy}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dy}{dx}\right)^2}\\ &= -\frac{\dfrac{dx}{dy}\cdot\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)}{\left(\dfrac{dy}{dx}\right)^2} \\ &= -\dfrac{\dfrac{dx}{dy}\cdot\dfrac{d^2y}{dx^2}}{\left(\dfrac{dy}{dx}\right)^2} \\ &= -\frac{\dfrac{d^2y}{dx^2}}{\left(\dfrac{dy}{dx}\right)^3}\end{aligned}$$

which is option 4.

Christopher Toni
  • 2,025
4

Assuming that all functions are defined properly, the inverse function theorem gives $$ \frac{dx}{dy} = \frac{1}{\frac{dy}{dx}}$$ Using the quotient rule and the chain rule you get $$ \frac{d}{dy}\left(\frac{dx}{dy}\right) =\frac{d}{dy}\left(\frac{1}{\frac{dy}{dx}}\right) =\frac{d}{dx}\left(\frac{1}{\frac{dy}{dx}}\right)\times \frac{dx}{dy}\\ = \frac{-\frac{d 1}{dx}\times \frac{dy}{dx}-1\times\frac{d^2 y}{dx^2}}{\left({\frac{dy}{dx}}\right)^2}\times\frac{dx}{dy} = \frac{-\frac{d^2 y}{dx^2}}{\left({\frac{dy}{dx}}\right)^2}\times\frac{dx}{dy} = \frac{-\frac{d^2 y}{dx^2}}{\left({\frac{dy}{dx}}\right)^2}\times\frac{1}{\frac{dy}{dx}} = -\frac{\frac{d^2 y}{dx^2}}{\left({\frac{dy}{dx}}\right)^3}$$

gammatester
  • 18,827
  • I deleted/corrected/undeleted the answer and added two steps. Juggling with just $y$ and $x$ is somewhat confusing. Much clearer is the e.g. http://math.stackexchange.com/a/249264/61216 – gammatester Nov 14 '13 at 09:38