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

How do you find the derivative of h(θ)=csc⁡θ+eθcot⁡θh\left( \theta \right) = \csc \theta + {e^\theta }\cot \theta ?

Explanation

Solution

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

Complete step by step solution:
We have,
h(θ)=csc⁡θ+eθcot⁡θh\left( \theta \right) = \csc \theta + {e^\theta }\cot \theta .
Now differentiate with respect to θ\theta .
ddθ(h(θ))=ddx(csc⁡θ+eθcot⁡θ)\dfrac{d}{{d\theta }}\left( {h\left( \theta \right)} \right) = \dfrac{d}{{dx}}\left( {\csc \theta + {e^\theta }\cot \theta } \right)
⇒h′(θ)=ddθ(csc⁡θ)+ddθ(eθcot⁡θ) −−(1)\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 ddθ(csc⁡θ)=−csc⁡θcot⁡θ −−−(2)\dfrac{d}{{d\theta }}\left( {\csc \theta } \right) = - \csc \theta \cot \theta {\text{ }} - - - (2) .
To find the differentiation of ddθ(eθcot⁡θ)\dfrac{d}{{d\theta }}\left( {{e^\theta }\cot \theta } \right) we apply product law.
ddθ(eθcot⁡θ)=eθddθ(cot⁡θ)+cot⁡θddθ(eθ)\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 ddθ(cot⁡θ)=−csc⁡2θ\dfrac{d}{{d\theta }}\left( {\cot \theta } \right) = - {\csc ^2}\theta and ddθ(eθ)=eθ\dfrac{d}{{d\theta }}({e^\theta }) = {e^\theta } .
ddθ(eθcot⁡θ)=−eθ.csc⁡2θ+cot⁡θ.eθ −−−(3)\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,
⇒h′(θ)=−csc⁡θcot⁡θ−eθ.csc⁡2θ+cot⁡θ.eθ\Rightarrow h'\left( \theta \right) = - \csc \theta \cot \theta - {e^\theta }.{\csc ^2}\theta + \cot \theta .{e^\theta }
⇒h′(θ)=−csc⁡θcot⁡θ−eθ(csc⁡2θ−cot⁡θ)\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′(θ)=−csc⁡θcot⁡θ−eθ(csc⁡2θ−cot⁡θ)h'\left( \theta \right) = - \csc \theta \cot \theta - {e^\theta }({\csc ^2}\theta - \cot \theta ) ”.

Note : We know the differentiation of xn{x^n} is d(xn)dx=n.xn−1\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.