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I'm trying to find $y''$ by implicit differentiation of this problem: $4x^2 + y^2 = 3$

So far, I was able to get $y'$ which is $\frac{-4x}{y}$

How do I go about getting $y''$? I am kind of lost on that part.

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You have $$y'=-\frac{4x}y\;.$$ Differentiate both sides with respect to $x$:

$$y''=-\frac{4y-4xy'}{y^2}=\frac{4xy'-4y}{y^2}\;.$$

Finally, substitute the known value of $y'$:

$$y''=\frac4{y^2}\left(x\left(-\frac{4x}y\right)-y\right)=-\frac4{y^2}\cdot\frac{4x^2+y^2}y=-\frac{4(4x^2+y^2)}{y^3}\;.$$

But from the original equation we know that $4x^2+y^2=3$, so in the end we have

$$y''=-\frac{12}{y^3}\;.$$

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    Shouldnt the answer be $\frac{-12}{y^3}$2012-06-29
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    Since $4x^2+y^2$ since x and y must satisfy the original equation2012-06-29
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    @soniccool: Yes: I inadvertently dropped a factor of $4$ in one term near the end. It’s fixed now.2012-06-29
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    Yes i was right!! Im learning!2012-06-29