How do you find the derivative of $h (θ) = csc θ + e^{θ} cot θ$ $h(theta) = csctheta +e^thetacottheta$ ?

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Answer 1 · 2018-06-01T13:18:05

$h'(theta)=(e^theta*cos(theta)*sin(theta)-cos(theta)-e^(theta))/sin^2(theta)$

We have
$h'(theta)=-csc(theta)cot(theta)+e^thetacot(theta)+e^theta(-1-cot^2(theta))$

Writing
$csc(theta)=1/sin(theta)$
$cot(theta)=cos(theta)/sin(theta)$
we get the result above.

How do you find the derivative of h(theta) = csctheta +e^thetacotthetah(θ)=cscθ+eθcotθh(theta) = csctheta +e^thetacottheta?