To find the derivative of the function [tex]\( f(u) = e^u \cot(u) \)[/tex], we will use the product rule and the chain rule. The product rule states that if we have two functions [tex]\( g(u) \)[/tex] and [tex]\( h(u) \)[/tex], then the derivative of their product is given by:
[tex]\[
(f \cdot g)' = f' \cdot g + f \cdot g'
\][/tex]
In this case, let [tex]\( f(u) = e^u \)[/tex] and [tex]\( g(u) = \cot(u) \)[/tex]. Then the product rule will be applied as follows:
1. Differentiate [tex]\( f(u) \)[/tex]:
[tex]\[
f'(u) = \frac{d}{du} e^u = e^u
\][/tex]
2. Differentiate [tex]\( g(u) \)[/tex]:
[tex]\[
g'(u) = \frac{d}{du} \cot(u)
\][/tex]
To find [tex]\( \frac{d}{du} \cot(u) \)[/tex], we use the fact that:
[tex]\[
\cot(u) = \frac{\cos(u)}{\sin(u)}
\][/tex]
Using the quotient rule for derivatives [tex]\(\left( \frac{f}{g} \right)' = \frac{f'g - fg'}{g^2} \)[/tex] where [tex]\( f(u) = \cos(u) \)[/tex] and [tex]\( g(u) = \sin(u) \)[/tex], we get:
[tex]\[
\frac{d}{du} \cot(u) = \frac{ (\cos(u))' \sin(u) - \cos(u) (\sin(u))' }{ (\sin(u))^2 }
\][/tex]
[tex]\[
= \frac{ -\sin(u) \sin(u) - \cos(u) \cos(u) }{ (\sin(u))^2 }
\][/tex]
[tex]\[
= \frac{ -(\sin^2(u) + \cos^2(u)) }{ (\sin(u))^2 }
\][/tex]
Using the Pythagorean identity [tex]\( \sin^2(u) + \cos^2(u) = 1 \)[/tex]:
[tex]\[
= \frac{ -1 }{ (\sin(u))^2 }
\][/tex]
Since [tex]\( \cot(u) = \frac{1}{\tan(u)} \)[/tex] and [tex]\(\frac{1}{\sin^2(u)} = \csc^2(u)\)[/tex], we have:
[tex]\[
\frac{d}{du} \cot(u) = -\csc^2(u)
\][/tex]
3. Apply the product rule. Now we have:
[tex]\[
\frac{d}{du} (e^u \cot(u)) = e^u \cdot \frac{d}{du} \cot(u) + \cot(u) \cdot \frac{d}{du} e^u
\][/tex]
[tex]\[
= e^u \cdot (-\csc^2(u)) + \cot(u) \cdot e^u
\][/tex]
So we get:
[tex]\[
\frac{d}{du} (e^u \cot(u)) = -e^u \csc^2(u) + e^u \cot(u)
\][/tex]
Or, in a simplified form using that [tex]\(\csc^2(u) = 1 + \cot^2(u)\)[/tex]:
[tex]\[
= (-\cot^2(u) - 1) e^u + e^u \cot(u)
\][/tex]
Thus, the derivative of the function [tex]\( f(u) = e^u \cot(u) \)[/tex] is:
[tex]\[
\boxed{(-\cot^2(u) - 1)e^u + e^u \cot(u)}
\][/tex]
