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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1 Answer
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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
$$y''=-\frac{12}{y^3}\;.$$
answered Jun 29, 2012 at 20:48
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$\begingroup$ Shouldnt the answer be $\frac{-12}{y^3}$ $\endgroup$ Commented Jun 29, 2012 at 20:49
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$\begingroup$ Since $4x^2+y^2$ since x and y must satisfy the original equation $\endgroup$ Commented Jun 29, 2012 at 20:56
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$\begingroup$ @soniccool: Yes: I inadvertently dropped a factor of $4$ in one term near the end. It’s fixed now. $\endgroup$ Commented Jun 29, 2012 at 21:00
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3$\begingroup$ Yes i was right!! Im learning! $\endgroup$ Commented Jun 29, 2012 at 21:01