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.

Can somebody tell me if I'm right on this? The math looks right, yet it just feels so wrong due to the obscene steps I had to take to get it.

I hope I transcribed all that correctly from my paper.

  1. $y=\tan(xy^3)$
  2. $y'=\sec^2(xy^3)(y^3+3xy^2y')$
  3. $\frac{y'}{\sec^2(xy^3)}=y^3+3xy^2y'$
  4. $\frac{y'}{\sec^2(xy^3)}-3xy^2y'=y^3$
  5. $y'(\cos^2(xy^3)-3xy^2)=y^3$
  6. $y'=\frac{y^3}{\cos^2(xy^3)}$

Edited to substitute $\frac{1}{\sec^2(xy^3)}$ for $\cos^2(xy^3)$

share|improve this question

1 Answer 1

up vote 2 down vote accepted

Looks good, but I'd rewrite it after step $5$ using $$\frac1{\sec^2(xy^3)}=\cos^2(xy^3).$$

In response to OP's edit: Now you've goofed going from $5$ to $6$. Take another look.

share|improve this answer
    
Good point. I absolutely didn't think of that. Thanks –  agent154 Oct 26 '12 at 2:23
    
Check my updated answer. –  Cameron Buie Oct 26 '12 at 20:03

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.