Question

How do you find the derivative for $f\left( x \right)=\dfrac{\cot x}{\sin x}$?

Answer 1

In this problem they have asked to calculate the derivative of the given function. We can observe that the given function is a fraction with the trigonometric ratios as both numerator and denominator. In differentiation we have the $\dfrac{u}{v}$ formula which is $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$. For applying this formula, we have to calculate the values of $\dfrac{du}{dx}$, $\dfrac{dv}{dx}$. So, we will compare the given equation with $\dfrac{u}{v}$ and calculate the values of $\dfrac{du}{dx}$, $\dfrac{dv}{dx}$. After calculating the values of $\dfrac{du}{dx}$, $\dfrac{dv}{dx}$ we will substitute them in the $\dfrac{u}{v}$ formula and simplify the obtained equation to get the required result.

Complete step by step answer:
Given function, $f\left( x \right)=\dfrac{\cot x}{\sin x}$.
Comparing the above function with $\dfrac{u}{v}$ form, then we will get
$u=\cot x$, $v=\sin x$.
Differentiating the both the above equation with respect to $x$, then we will get
$\dfrac{du}{dx}=\dfrac{d}{dx}\left( \cot x \right)$, $\dfrac{dv}{dx}=\dfrac{d}{dx}\left( \sin x \right)$.
We have the differentiation formulas $\dfrac{d}{dx}\left( \cot x \right)=-{{\csc }^{2}}x$, $\dfrac{d}{dx}\left( \sin x \right)=\cos x$. Substituting these values in the above equation, then we will get
$\Rightarrow \dfrac{du}{dx}=-{{\csc }^{2}}x$, $\Rightarrow \dfrac{dv}{dx}=\cos x$.
Now differentiating the given function with respect to $x$, then we will get
${{f}^{'}}\left( x \right)=\dfrac{d}{dx}\left( \dfrac{\cot x}{\sin x} \right)$
Applying the differentiation formula $\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}$ in the above equation, then we will get
$\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{\sin x\left( \dfrac{du}{dx} \right)-\cot x\left( \dfrac{dv}{dx} \right)}{{{\sin }^{2}}x}$
Substituting the values of $\dfrac{du}{dx}$, $\dfrac{dv}{dx}$ in the above equation, then we will get
$\Rightarrow {{f}^{'}}\left( x \right)=\dfrac{\sin x\left( -{{\csc }^{2}}x \right)-\cot x\cos x}{{{\sin }^{2}}x}$
Simplifying the above equation by converting the all-trigonometric ratios into $\sin x$, $\cos x$, then we will get
$\begin{aligned} & \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{-\sin x\times \dfrac{1}{{{\sin }^{2}}x}-\dfrac{\cos x}{\sin x}\times \cos x}{{{\sin }^{2}}x} \\\ & \Rightarrow {{f}^{'}}\left( x \right)=\dfrac{\dfrac{-1}{\sin x}-\dfrac{{{\cos }^{2}}x}{\sin x}}{{{\sin }^{2}}x} \\\ & \Rightarrow {{f}^{'}}\left( x \right)=-\dfrac{1+{{\cos }^{2}}x}{{{\sin }^{2}}x\times \sin x} \\\ & \Rightarrow {{f}^{'}}\left( x \right)=-\dfrac{1+{{\cos }^{2}}x}{{{\sin }^{3}}x} \\\ \end{aligned}$

Hence the derivative of the given function $f\left( x \right)=\dfrac{\cot x}{\sin x}$ is $-\dfrac{1+{{\cos }^{2}}x}{{{\sin }^{3}}x}$.

Note: In this problem we have converted all the trigonometric ratios into $\sin x$, $\cos x$ after calculating the derivative of the given function. We can also simplify the given equation by substituting $\cot x=\dfrac{\cos x}{\sin x}$ in the given function and simplify the function. Now we will get the whole given equation in terms of $\sin x$, $\cos x$. Now we can simplify derivative.