# Implicit differentiation with sin functions

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 $y+x\cos(y) = x^2y$

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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
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. 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 $$y^\prime = \frac{2xy - \cos(y)}{1 - x\sin(y) - x^2}.$$