Question

Differentiate
\[h(\theta ) = 5\csc \theta +2e^\theta \cot \theta\]

Answer 1

not sure how i can compute the csc and cot

Answer 2

write them in terms of sin and cos

Answer 3

csc = -csccot

cot = -csc^2

Answer 4

\[-5csc(t)cot(t)+2t.e^t cot(t)-2e^t csc(t)\]

Answer 5

should be a ^2 on the last csc

Answer 6

\[....-2e^tcsc^2(t)\]

Answer 7

\[\frac{d}{d \theta}(h(\theta) = \frac{d}{d \theta}[2 e \theta \cot(\theta)+5 \csc(\theta)]\]\[\rightarrow\ [h'(\theta) = \frac{d}{d \theta} [2 e \theta \cot(\theta)+5 \csc(\theta)]\]\[\rightarrow h'(\theta) = 2 e \frac{d}{d \theta}[\theta \cot(\theta)]+5 \frac {d}{d \theta}[\csc(\theta)]\]\[\rightarrow h'(\theta) = 2 e [\cot(\theta) \frac{d}{d \theta}(\theta)+(\theta) \frac{d}{d \theta}\cot(\theta)+5 \frac{d}{d \theta}\csc(\theta)]\]\[\rightarrow h'(\theta) = 2 e [\theta \frac{d}{d \theta}\cot(\theta)+\cot(\theta) \frac{d}{d \theta}(\theta)+5 -(\cot(\theta) \csc(\theta))]\]\[\rightarrow h'(\theta) = 2 e (\theta \frac{d}{d \theta}[\cot(\theta)]+[1 \cot(\theta)]-5 \cot(\theta) \csc(\theta)\]\[\rightarrow h'(\theta) = 2 e [\cot(\theta)+\theta (-\csc^2(\theta)]-5 \cot(\theta) \csc(\theta)\]

Answer 8

nice work; but i believe the original stated \(2e^\theta\)

Answer 9

Ah yes, you're right...