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

Question

Emilia Aguilar · Accepted Answer

I would rewrite #sin^2xcos^2x #

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

#f(x) = sin^2xcos^2x = sin^2x(1-sin^2x) = sin^2x-sin^4x#

#f'(x) = 2sinxcosx-4sin^3xcosx#

Rewrite as needed.

Of course, we could have used #f(x) = cos^2x-cos^4x#

to get #f'(x) = -2cosxsinx+4cos^3xsinx#.

Employ a Double Angle Identity

#sin2x = 2sinxcosx# so

#f(x) = (sinxcosx)^2 = (1/2sin2x)^2 = 1/4 sin^2 2x#.

So, #f'(x) = 1/4[2sin2xd/dx(sin2x)]#

# = 1/2sin2x cos2x d/dx(2x)#

# = sin2x cos2x#

Edit if you'd like.

Ezra Adams · Answer

undefined 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) $.