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

Answer 1

Isaiah Allen

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

Apply the rule of the product:

#f'(x)=cos(2x)d/dx[e^(2x)]+e^(2x)d/dx[cos(2x)]#

Both of these derivatives need to be found separately, using the chain rule.

#d/dx[e^(2x)]=e^(2x)d/dx[2x]=2e^(2x)#

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

Re-plug these in.

#f'(x)=2e^(2x)cos(2x)-2e^(2x)sin(2x)#

#f'(x)=2e^(2x)(cos(2x)-sin(2x))#

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Answer 2

Ezekiel Alexander

To find the derivative of (e^{2x} \times \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 (u'(x)v(x) + u(x)v'(x)).

Let (u(x) = e^{2x}) and (v(x) = \cos(2x)). Then, find the derivatives of each function:

(u'(x) = 2e^{2x}) (derivative of (e^{2x}) with respect to (x))

(v'(x) = -2\sin(2x)) (derivative of (\cos(2x)) with respect to (x), using the chain rule)

Now, apply the product rule:

(u'(x)v(x) + u(x)v'(x) = (2e^{2x})\cos(2x) + e^{2x}(-2\sin(2x)))

So, the derivative of (e^{2x} \times \cos(2x)) is (2e^{2x}\cos(2x) - 2e^{2x}\sin(2x)).

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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.