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 can't find an example in my book so I am not sure how I am suppose to do this.

I am trying to find the derivative of $y$ for $y+x\cos(y) = x^2y$

share|improve this question
    
use chain rule. –  leo Oct 9 '11 at 22:32
    
So the xcosy is 1(cosy)+x(-siny)? –  user138246 Oct 9 '11 at 22:36
1  
Close. Don't say "is," because you're not saying they're equal. And you are differentiating with respect to $x$, so the derivative of $\cos y$ - using the chain rule - is $(-\sin y)y'$. And that's just differentiating the left-hand side; now try what's on the right. –  anon Oct 9 '11 at 22:42
    
use the chain and product rules. –  robjohn Oct 9 '11 at 22:42
    
I got $y\prime + cosy(y\prime) - 2xy\prime y$ –  user138246 Oct 9 '11 at 22:50

1 Answer 1

When you do implicit differentiation problems, there are three important things to keep in mind.

  1. Whenever you have a mixture of $x$ and $y$ factors, you must use the product/quotient rule.
  2. Whenever you differentiate a term involving $y$, you must include a factor of $y^\prime$ (since we are differentiating with respect to $x$, not $y$).
  3. If you did not differentiate a factor of $y$, you do not include a factor of $y^\prime$.

Keeping these things in mind, I get $$ 1 \cdot y^\prime + (\cos(y) - x\sin(y)y^\prime) = (2xy + x^2\cdot 1 \cdot y^\prime), $$ where I've included parentheses to show where product rule is taking place.

Now, the whole point of this business was to get $y^\prime$ by itself. So, move everything having to do with $y^\prime$ to one side of the equation and all other terms to the other. $$ y^\prime - x\sin(y)y^\prime - x^2y^\prime = 2xy - \cos(y). $$ Factoring out the $y^\prime$ gives $$ y^\prime(1 - x\sin(y) - x^2) = 2xy - \cos(y). $$ Finally, dividing to isolate $y^\prime$ leaves us with $$ y^\prime = \frac{2xy - \cos(y)}{1 - x\sin(y) - x^2}. $$

share|improve this answer

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.