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Complex Differentiation

Any help with this question would be highly appreciated,

Find the x-coordinates of the stationary points of
y=x^3 e^-kx, where k is a positive constant.

Thanks

Reply 1

user2020user

Have you encountered the product and chain rules of differentiation yet?

\dfrac{d}{dx} \left( u(x) \cdot v(x) \right) = u'(x) \cdot v(x) + u(x) \cdot v'(x)

(edited 10 years ago)

Reply 2

SimonHackland

Original post by Khallil

Have you encountered the product and chain rules of differentiation yet?

\dfrac{d}{dx} \left( u(x) \cdot v(x) \right) = u'(x) \cdot v(x) + u(x) \cdot v'(x)

I have used the product rule to get to this stage,

dy/dx=x^3(-ke^-kx)+e^-kx(3x^2)

I am unsure what to do from now onwards

Reply 3

user2020user

Original post by SimonHackland

I have used the product rule to get to this stage,

dy/dx=x^3(-ke^-kx)+e^-kx(3x^2)

I am unsure what to do from now onwards

Stationary points occur when the derivative of a function is 0.

You should also note that

\forall x : e^{kx} \neq 0

so you can factor that out and omit it since it doesn't have any real solutions.

Reply 4

SimonHackland

I have done it now, thanks for the help

(edited 10 years ago)

Reply 5

lubus

Original post by SimonHackland

I have done it know, thanks for the help

*now

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