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?
Find $y'$ if $\sin(x+y) = \cos(xy)$
1
$\begingroup$
implicit-differentiation
-
1It is fine now. – 2012-10-16
2 Answers
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 $y'$.
-
1Let'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. – 2012-10-16
1
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.