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How do you differentiate #y=e^(-5x)cos3x#?

Answer 1

Ezra Adams

Use the Product Rule

The product rule is:

#(d(gh))/dx = g'(h) + g(h')#

Let #g = e^(-5x) and h = cos(3x)#, then:

#g' = -5e^(-5x) and h' = -3sin(3x)#

Substituting into the product rule:

#(d(e^(-5x)cos(3x)))/dx = -5e^(-5x)cos(3x) -3e^(-5x)sin(3x)#

#(d(e^(-5x)cos(3x)))/dx = -e^(-5x)(5cos(3x) +3sin(3x))#

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Answer 2

Isabella Ackerman

To differentiate ( y = e^{-5x}\cos(3x) ), you would use the product rule. The product rule states that if you have two functions, ( u(x) ) and ( v(x) ), then the derivative of their product, ( u(x) \cdot v(x) ), is given by ( u'(x) \cdot v(x) + u(x) \cdot v'(x) ).

In this case, ( u(x) = e^{-5x} ) and ( v(x) = \cos(3x) ).

The derivatives of ( e^{-5x} ) and ( \cos(3x) ) with respect to ( x ) are ( -5e^{-5x} ) and ( -3\sin(3x) ), respectively.

Therefore, applying the product rule:

[ \frac{d}{dx}(e^{-5x}\cos(3x)) = -5e^{-5x}\cos(3x) + e^{-5x}(-3\sin(3x)) ]

[ = -5e^{-5x}\cos(3x) - 3e^{-5x}\sin(3x) ]

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Answer from HIX Tutor

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