Extreme Value Theory/Derivatives - Urgent - Please help me

[Jamez1988]

I am going through past papers for my exam on Tuesday and I am getting stuck trying to figure out how my lecturer came to the solution of this question.
The question is:
Given is the function f(x) = 5 + xe^-5x, with domain 0≤x≤1. Apply the extreme value theorem to the above function.
My lecturer has given the answer as: f'(x) = e^-5x + xe^-5 .-5 which he then simplifies to f'(x) = e^-5x(1-5x)
How did he come to this solution? Specifcally, whey does he multiply by -5 and how is the simplification (final answer) equal to the initial?

I have looked at every different definition/equation and just cannot work this out

 

[Deleted member 4993]

I am going through past papers for my exam on Tuesday and I am getting stuck trying to figure out how my lecturer came to the solution of this question.
The question is:
Given is the function f(x) = 5 + xe^-5x, with domain 0≤x≤1. Apply the extreme value theorem to the above function.
My lecturer has given the answer as: f'(x) = e^-5x + xe^-5 .-5 which he then simplifies to f'(x) = e^-5x(1-5x) [ ... this should not be in exponent]
How did he come to this solution? Specifcally, whey does he multiply by -5 and how is the simplification (final answer) equal to the initial?

I have looked at every different definition/equation and just cannot work this out

I am going to rewrite stuff that you wrote:

f'(x) = e^-5x + xe^-5x * (-5) .... you missed that x in the second part. Now factor out e^-5x

You get:

f'(x) = e^-5x + xe^-5x * (-5)

= e^-5x * (1 - 5x) [or to avoid confusion with exponent]

= (1 - 5x) * e^-5x

 

[Jamez1988]

I am going to rewrite stuff that you wrote:

f'(x) = e^-5x + xe^-5x * (-5) .... you missed that x in the second part. Now factor out e^-5x

You get:

f'(x) = e^-5x + xe^-5x * (-5)

= e^-5x * (1 - 5x) [or to avoid confusion with exponent]

= (1 - 5x) * e^-5x

Thanks - The bit really confusing me, is why do we * the (-5)?

 

[HallsofIvy]

That's the "chain rule". If y= f(u(x)) so that y is a function of u and u is a function of x, \(\displaystyle \frac{dy}{dx}= \frac{dy}{du}\frac{du}{dx}\).

In this problem, \(\displaystyle y= e^{-5x}\), we can take \(\displaystyle u= -5x\) so that \(\displaystyle y= e^u\). So \(\displaystyle \frac{dy}{du}= e^u\) and \(\displaystyle \frac{du}{dx}= -5\).

So \(\displaystyle \frac{dy}{dx}= \frac{dy}{du}\frac{du}{dx}= (e^u)(-5)= -5e^u= -5e^{-5x}\).