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?
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$\begingroup$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 $y'$.
$\endgroup$ 2 $\begingroup$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.
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