Question

How do you find the derivative of \[h\left( \theta \right) = \csc \theta + {e^\theta }\cot \theta \] ?

Answer 1

Hint: Derivatives are defined as the varying rate of a function with respect to an independent variable. Here we have a function of \[\theta \] . We need to differentiate the given equation with respect to \[\theta \] . We use product rule to solve this that is \[\dfrac{{dy}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}\] .

Complete step by step solution:
We have,
 \[h\left( \theta \right) = \csc \theta + {e^\theta }\cot \theta \] .
Now differentiate with respect to \[\theta \] .
 \[\dfrac{d}{{d\theta }}\left( {h\left( \theta \right)} \right) = \dfrac{d}{{dx}}\left( {\csc \theta + {e^\theta }\cot \theta } \right)\]
 \[\Rightarrow h'\left( \theta \right) = \dfrac{d}{{d\theta }}\left( {\csc \theta } \right) + \dfrac{d}{{d\theta }}\left( {{e^\theta }\cot \theta } \right){\text{ }} - - (1)\]
We know the \[\dfrac{d}{{d\theta }}\left( {\csc \theta } \right) = - \csc \theta \cot \theta {\text{ }} - - - (2)\] .
To find the differentiation of \[\dfrac{d}{{d\theta }}\left( {{e^\theta }\cot \theta } \right)\] we apply product law.
 \[\dfrac{d}{{d\theta }}\left( {{e^\theta }\cot \theta } \right) = {e^\theta }\dfrac{d}{{d\theta }}\left( {\cot \theta } \right) + \cot \theta \dfrac{d}{{d\theta }}({e^\theta })\]
We know \[\dfrac{d}{{d\theta }}\left( {\cot \theta } \right) = - {\csc ^2}\theta \] and \[\dfrac{d}{{d\theta }}({e^\theta }) = {e^\theta }\] .
 \[\dfrac{d}{{d\theta }}\left( {{e^\theta }\cot \theta } \right) = - {e^\theta }.{\csc ^2}\theta + \cot \theta .{e^\theta }{\text{ }} - - - (3)\]
Substituting equation (2) and (3) in equation (1) we have,
 \[\Rightarrow h'\left( \theta \right) = - \csc \theta \cot \theta - {e^\theta }.{\csc ^2}\theta + \cot \theta .{e^\theta }\]
 \[\Rightarrow h'\left( \theta \right) = - \csc \theta \cot \theta - {e^\theta }({\csc ^2}\theta - \cot \theta )\] , This is the required answer.
So, the correct answer is “ \[h'\left( \theta \right) = - \csc \theta \cot \theta - {e^\theta }({\csc ^2}\theta - \cot \theta )\] ”.

Note: We know the differentiation of \[{x^n}\] is \[\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}\] . The obtained result is the first derivative. If we differentiate again we get a second derivative. If we differentiate the second derivative again we get a third derivative and so on. Careful in applying the product rule. We also know that differentiation of constant terms is zero.