Question

Can some one help me find the derivative for this. I did it, but now I'm stuck

Answer 1

\[e ^\theta [\frac{ 1 }{ \theta^2 } + \theta ^ \frac{ -\pi }{ 2 }]\]

Answer 2

Well, might as well do product rule.
f(x) = e^(theta)
g(x) = ((1/(theta^2) + theta^(-pi/2))
f'(x) = e^(theta)

as for g'(x), the first term is just a basic power rule.
\[\frac{ 1 }{ \theta ^{2} }= \theta^{-2} \]The derivative of this portion then just becomes:
\[\frac{-2}{\theta^{3}}\]As for the second portion, its the same thing, just power rule.
\[\theta ^{- \pi/2} \implies \frac{- \pi}{2} \theta ^{\frac{ -\pi }{ 2 }-1}\]So we can say all of g'(x) is:
\[(-\frac{2}{\theta^{3}}- \frac{ \pi }{ 2 }*\theta ^{-\pi/2 - 1})\]
Then this can all be plugged into the product rule formula:

\[e^ {\theta}* ( \frac{ 1 }{\theta ^{2} }+\theta ^{\frac{ -\pi }{ 2 }})- e^{\theta} * (\frac{ 2 }{ \theta ^{3} }+\frac{ \pi }{ 2 }\theta ^{\frac{ -\pi }{ 2 }-1})\]
Its a bit of a mess, but long as it makes sense. Do you need to simplify this answer, too, or is this okay?

Answer 3

thats what i got, but idk if i should simplify it?