Find $y'$ if $\sin(x+y) = \cos(xy)$

Question

Find $y'$ if $\sin(x+y) = \cos(xy)$. I got ${-\cos(x+y)\over \sin(xy) * (x+y)}$. Please use Implicit Differentiation. Did I get the answer right?

ok ${-ysin(xy)-cos(x+y)\over cos(x+y)+xsin(xy)}$ correct? – dsta Oct 16 '12 at 05:05 — dsta, Oct 16 '12 at 05:05

score 2 · Answer 1 · answered Oct 16 '12 at 04:52

2

Applying $\frac{d}{dx}$ to both sides of $\sin(x+y)=\cos(xy)$ we get (using $y'$ as shorthand for $\frac{dy}{dx}$) $$(y'+1)\cos(x+y)=-(y+xy')\sin(xy)\,.$$

Now solve for $y'$.

answered Oct 16 '12 at 04:52

Bey

3,893

that's what I got but I don't know how to get the y' together – dsta Oct 16 '12 at 04:53
1

Let's just look at the left-hand side. Distribute to get $(y'+1)\cos(x+y) = y'\cos(x+y)+\cos(x+y)$. Distribute on the right-hand side, too. There will be two terms with a $y'$ in them; bring them both over to the same side of the equation (doesn't matter which), and move everything else to the opposite side. Factor out a $y'$, and divide. – Bey Oct 16 '12 at 04:55

André Nicolas · Answer 2 · 2012-10-16T04:56:49.713

Differentiate. By the Chain Rule, the derivative of $\sin(x+y)$ with respect to $x$ is $$\left(1+y'\right)\cos(x+y).$$

Similarly, the derivative of $\cos(xy)$ with respect to $x$ is $$\left(xy'+y\right)(-\sin(xy)).$$

The two expressions are identically equal. When you use them to find $y'$, you will get an expression which is different from the one you obtained.

Bring the $y'$ stuff to one side, and the rest to the other. We get $$y'(cos(x+y)+x\sin(xy))=-(\cos(x+y)+y\sin(xy)).$$ Finally, divide.

Find $y'$ if $\sin(x+y) = \cos(xy)$

2 Answers2