How do you differentiate #cos(3x)sin(3x)#?

Answer 1

See below.

You could use the product rule and the chain rule, but I find it more satisfying to rewrite the function before differentiating.

From trigonometry:

#sin(2theta) = 2sin(theta)cos(theta)#

Therefore

#sin(theta)cos(theta) = 1/2sin(2theta)#

So,

#f(x) = cos(3x)sin(3x) = 1/2sin(6x)#

Therefore,

#f'(x) = 1/2cos(6x) * d/dx(6x)# #" "# (chain rule)
# = 3cos(6x)#
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Answer 2

To differentiate the product cos(3x)sin(3x), you can use the product rule from calculus. The product rule states that if you have two functions, u(x) and v(x), then the derivative of their product is given by:

(uv)' = u'v + uv'

For the given functions cos(3x) and sin(3x), their derivatives are:

(cos(3x))' = -3sin(3x) (sin(3x))' = 3cos(3x)

Now, applying the product rule:

(cos(3x)sin(3x))' = (-3sin(3x))sin(3x) + cos(3x)(3cos(3x))

Simplify:

(cos(3x)sin(3x))' = -3sin^2(3x) + 3cos^2(3x)

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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.

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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