How do you find the derivative of #h(theta) = csctheta +e^thetacottheta#?

1 Answer
Sonnhard
Jun 1, 2018

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

Explanation:

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.

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