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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. $x$ $$2xy + x^2y' = 0$$$$ y' = \frac{-2xy}{x^2} = \frac{-2y}{x} $$


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

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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 $$\dfrac{dy}{dx} = -\dfrac2{x^3} \implies \dfrac{d^2y}{dx^2} = \dfrac6{x^4}$$

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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. – S. Snape Feb 3 '13 at 7:12

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