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 a slight problem with this: Given the equation $$ x^2y = 1$$

and asked to find $y''$, I attempted to apply implicit differentiation by differentiation w.r.t. $y$. $$2xy'y + x^2y' = 0$$

However, it does not seem to be right

UPDATE Solved: Differentiate w.r.t. $x$ $$2xy + x^2y' = 0$$$$ y' = \frac{-2xy}{x^2} = \frac{-2y}{x} $$


$$y'' =\frac{x(-2y') - (-2y)(1)}{x^2} = \frac{-2xy' + 2y}{x^2} = \frac{6y}{x^2}$$

share|cite|improve this question
In order to find $y''$, you should be differentiating implicitly with respect to $x$, not $y$. (And of course you’ll have to get from $y'$ to $y''$ as well.) Your first differentiation should result in $2xy+x^2y'=0$. – Brian M. Scott Feb 3 '13 at 6:50
up vote 4 down vote accepted

We are given that $x^2y = 1 \,\,\,\,\, (\spadesuit)$.

Hence, we have $$\dfrac{d(x^2y)}{dx} = 0 \implies \dfrac{d(x^2)}{dx}y + x^2 \dfrac{dy}{dx} = 0\implies 2xy + x^2 \dfrac{dy}{dx} = 0 \implies 2y + x\dfrac{dy}{dx}=0 \,\,\, (\star)$$ Now differentiate again to get $$\dfrac{d}{dx} \left(2y + x \dfrac{dy}{dx}\right) = 0 \implies 2 \dfrac{dy}{dx} + \dfrac{d}{dx} \left(x \dfrac{dy}{dx}\right) = 0\\ \implies 2 \dfrac{dy}{dx} + \dfrac{dx}{dx} \dfrac{dy}{dx} + x \dfrac{d^2y}{dx^2} = 0 \implies 3 \dfrac{dy}{dx} + x \dfrac{d^2y}{dx^2} = 0 \,\,\,\, (\dagger)$$ From $(\star)$, we have $\dfrac{dy}{dx} = - \dfrac{2y}x$. Plugging this in $(\dagger)$, we get that $$x \dfrac{d^2y}{dx^2} + 3 \times \left(- \dfrac{2y}x\right) = 0 \implies \dfrac{d^2y}{dx^2} = \dfrac{6y}{x^2}$$ From $(\spadesuit)$, we have $y = \dfrac1{x^2}$ and hence $$\dfrac{d^2y}{dx^2} = \dfrac6{x^4}$$ You could do a direct differentiation by noticing that $y = \dfrac1{x^2}$. This implies $$\dfrac{dy}{dx} = -\dfrac2{x^3} \implies \dfrac{d^2y}{dx^2} = \dfrac6{x^4}$$

share|cite|improve this answer
Okay, from today on, I'll just remember that $y'$ is $\frac {dy}{dx}$ and remember to choose the right variable to differentiate w.r.t. to – bryansis2010 Feb 3 '13 at 7:07
@bryansis2010: Assuming $F(x,y)=0$ defines $y$ recpect to $x$ as a function implicity. Thereofre, it can be easily proved that $$y'=-\frac{F_x}{F_y}$$ as well. – Babak S. Feb 3 '13 at 7:12

Your Answer


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

Not the answer you're looking for? Browse other questions tagged or ask your own question.