# How do you find the derivative of #sin^2(x)cos^2(x)#?

I would rewrite

I would prefer one of the two below, but we could use the product, power, and chain rules as well:

Use the identity of Pythagorean

Rewrite as needed.

Employ a Double Angle Identity

Edit if you'd like.

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To find the derivative of ( \sin^2(x) \cos^2(x) ), you can use the product rule of differentiation. The product rule states that if you have two functions, ( u(x) ) and ( v(x) ), the derivative of their product ( u(x) \cdot v(x) ) with respect to ( x ) is given by ( u'(x) \cdot v(x) + u(x) \cdot v'(x) ).

Let ( u(x) = \sin^2(x) ) and ( v(x) = \cos^2(x) ). Then, ( u'(x) ) is the derivative of ( \sin^2(x) ) with respect to ( x ), and ( v'(x) ) is the derivative of ( \cos^2(x) ) with respect to ( x ).

Using the chain rule, the derivatives are:

[ u'(x) = 2 \sin(x) \cos(x) ] [ v'(x) = -2 \sin(x) \cos(x) ]

Now, apply the product rule:

[ \frac{d}{dx} (\sin^2(x) \cos^2(x)) = u'(x) \cdot v(x) + u(x) \cdot v'(x) ] [ = (2 \sin(x) \cos(x)) \cdot (\cos^2(x)) + (\sin^2(x)) \cdot (-2 \sin(x) \cos(x)) ] [ = 2 \sin(x) \cos(x) \cos^2(x) - 2 \sin(x) \cos(x) \sin^2(x) ]

So, the derivative of ( \sin^2(x) \cos^2(x) ) is ( 2 \sin(x) \cos(x) \cos^2(x) - 2 \sin(x) \cos(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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