Step 1:

Notice that this is an implicit function, where the dependent variable y appears on both sides. To find the derivative of an implicit function, we will use implicit differentiation.

First, we find the derivative of both sides with respect to x, as follows:

[latex]\large{\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2)  = \frac{d}{dx}(\sin(xy))}[/latex]

Step 2:

Let’s now individually evaluate each term. Remember, when dealing with implicit differentiation, treat the y as y(x) and this will help us when differentiating.

[latex]\large{\frac{d}{dx}(x^2) = 2x}[/latex]

[latex]\large{\frac{d}{dx}(y^2) = 2yy’}[/latex]

[latex]\large{\frac{d}{dx}(\sin(xy)) = \cos(xy) \cdot \frac{d}{dx}(xy)}[/latex]

[latex]\large{ = \cos(xy) \cdot (xy’ + y)}[/latex]

[latex]\large{= xy’\cos(xy) + y\cos(xy)}[/latex]

Note that we needed multiple steps to find [latex]\frac{d}{dx}\sin(xy)[/latex] since we had to apply the chain rule.

Step 3:

Now, combining all the terms, we have:

[latex]\large{2x +2yy’ = xy’\cos(xy)+y\cos(xy)}[/latex]

Remember, the goal is to find the derivative, or [latex]y'[/latex]. Thus, we must isolate for [latex]y'[/latex]. First, we will bring all terms with [latex]y'[/latex] to one side as follows:

[latex]\large{2yy’ -xy’\cos(xy) = y\cos(xy) – 2x}[/latex]

Step 4:

Now, we will factor out [latex]y'[/latex] and solve for it, as follows:

[latex]\large{y'(2y – x\cos(xy)) = y\cos(xy) – 2x}[/latex]

[latex]\large{y’ = \frac{y\cos(xy)-2x}{2y-x\cos(xy)}}[/latex]

Thus, we have found the derivative:

[latex]\large{y’ = \frac{dy}{dx} = \frac{y\cos(xy)-2x}{2y-x\cos(xy)}}[/latex]

Solution Complete!