# What is the derivative of #f(x) = sin^2(x)+xcos^2(x)#?

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To find the derivative of ( f(x) = \sin^2(x) + x\cos^2(x) ), we differentiate each term separately using the chain rule and product rule where necessary.

( f'(x) = \frac{d}{dx}(\sin^2(x)) + \frac{d}{dx}(x\cos^2(x)) )

Using the chain rule, (\frac{d}{dx}(\sin^2(x)) = 2\sin(x)\cos(x)).

For ( \frac{d}{dx}(x\cos^2(x)) ), we apply the product rule, resulting in ( \cos^2(x) - 2x\sin(x)\cos(x) ).

So, putting it all together, the derivative of ( f(x) ) is ( f'(x) = 2\sin(x)\cos(x) + \cos^2(x) - 2x\sin(x)\cos(x) ). Simplifying, we get ( f'(x) = \cos^2(x) + 2\sin(x)\cos(x) - 2x\sin(x)\cos(x) ).

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The derivative of (f(x) = \sin^2(x) + x\cos^2(x)) is (f'(x) = 2\sin(x)\cos(x) + \cos^2(x) - x\sin^2(x)).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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