Home > Calculus > Derivative Rules for y=cos(x) and y=tan(x)

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

Answer 1

Emilia Aguilar

Method 1

Apply the chain and product rules.

#d/dx(sin(2x)cos(2x)) = d/dx(sin(2x))cos(2x)+sin(2x)d/dx(cos(2x))#

# = [cos(2x)d/dx(2x)]cos(2x)+sin(2x)[-sin(2x)d/dx(2x)]#

# = 2cos^2(2x)-2sin^2(2x)#

This can be rewritten using trigonometry.

Method 2

Use #sin(2theta) = 2sintheta cos theta# to write

#sin(2x)cos(2x)=1/2sin(4x)#

Apply the chain rule now.

#d/dx (1/2sin(4x)) = 1/2 cos(4x)d/dx(4x)#

# = 1/2 cos(4x)*4 = 2cos(4x)#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Charles Arocha

To find the derivative of sin(2x)cos(2x), you can use the product rule of differentiation. The product rule states that if you have two functions u(x) and v(x), then the derivative of their product is given by the formula: (u'v + uv'). Applying this rule to sin(2x)cos(2x), where u(x) = sin(2x) and v(x) = cos(2x), we get:

u'(x) = d/dx(sin(2x)) = 2cos(2x) v'(x) = d/dx(cos(2x)) = -2sin(2x)

Now, using the product rule, we have:

(sin(2x)cos(2x))' = (2cos(2x))cos(2x) + sin(2x)(-2sin(2x))

Simplify this expression to get the derivative of sin(2x)cos(2x).

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

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.