Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 3 down vote accepted

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

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

share|improve this answer
    
Shouldnt the answer be $\frac{-12}{y^3}$ –  soniccool Jun 29 '12 at 20:49
    
Since $4x^2+y^2$ since x and y must satisfy the original equation –  soniccool Jun 29 '12 at 20:56
    
@soniccool: Yes: I inadvertently dropped a factor of $4$ in one term near the end. It’s fixed now. –  Brian M. Scott Jun 29 '12 at 21:00
3  
Yes i was right!! Im learning! –  soniccool Jun 29 '12 at 21:01

Your Answer

 
discard

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.