# How do you differentiate #y^2-x^3cos(x) = x^2-y#?

Using the chain and the product rule we get

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To differentiate the equation (y^2 - x^3 \cos(x) = x^2 - y), you would use implicit differentiation.

Taking the derivative of each term with respect to (x), you get:

[\frac{d}{dx} (y^2) - \frac{d}{dx} (x^3 \cos(x)) = \frac{d}{dx} (x^2) - \frac{d}{dx} (y)]

Using the chain rule for the second term on the left side, you have:

[2y \frac{dy}{dx} - (3x^2 \cos(x) - x^3 \sin(x)) = 2x - \frac{dy}{dx}]

Now, solve for (\frac{dy}{dx}):

[2y \frac{dy}{dx} + \frac{dy}{dx} = 3x^2 \cos(x) - x^3 \sin(x) + 2x - y]

[3y \frac{dy}{dx} = 3x^2 \cos(x) - x^3 \sin(x) + 2x - y]

[ \frac{dy}{dx} = \frac{3x^2 \cos(x) - x^3 \sin(x) + 2x - y}{3y}]

That's the derivative of the given equation with respect to (x).

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