Question

If sin(xy) = x^2, then dy/dx= ?

Show all work

Answer 1

Answer:

Explanation:

$\sin(xy)=x^2$

We want to differentiate both sides since we are to find $\frac{dy}{dx}$.

$(\sin(xy))'=(x^2)'$

The right hand side is easy and is one step. You apply the power rule on the right hand side. The side that is going to be a little hard is the left hand side.

So let's focus on that for now.

$(\sin(xy))'$

We will begin with chain rule:

$(xy)'\cos(xy)$

Now we must use the product rule to differentiate the $(xy)$ part:

$(xy'+1y)\cos(xy)$

$(xy'+y)\cos(xy)$

Distribute:

$xy'\cos(xy)+y\cos(xy)$

So we are done differentiate the left hand side.

The right hand side after applying power rule gives us $(x^2)'=2x$.

Let's put it together:

$xy'\cos(xy)+y\cos(xy)=2x$

We need to get the term(s) with $y'$ alone. There is only one of these luckily.  So moving the one term of $y\cos(xy)$ to the other side will give us victory in getting the term with $y'$ alone.

$xy'\cos(xy)+y\cos(xy)=2x$

Subtracting $y\cos(xy)$ on both sides gives:

$xy'\cos(xy)=2x-y\cos(xy)$

Dividing both sides by $x\cos(xy)$ gives:

$y'=\frac{2x-y\cos(xy)}{x\cos(xy)}$

We could write $y'$ as $\frac{dy}{dx}$ giving us:

$\frac{dy}{dx}=\frac{2x-y\cos(xy)}{x\cos(xy)}$