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

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Answer:

Step-by-step explanation:


\sin(xy)=x^2

We want to differentiate both sides since we are to find
(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'=(2x-y\cos(xy))/(x\cos(xy))

We could write
y' as
(dy)/(dx) giving us:


(dy)/(dx)=(2x-y\cos(xy))/(x\cos(xy))

User Jeeyoung
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