Question

Solve the differentiation $\dfrac{d}{dx}(\dfrac{{{\cot }^{2}}x-1}{{{\cot }^{2}}x+1})=$.

Answer 1

Hint: We will use a simple concept of trigonometry and convert $\cot x$ in terms of $\sin x$ and $\cos x$ to make the question easier than before. We have to use formula, $\cot x=\dfrac{\cos x}{\sin x}$ then, differentiate and easily get the answer.

Complete step by step solution:
If the question of differentiation contains trigonometric terms, then first of all we have to simplify the trigonometric term using trigonometric formula and trigonometric identities which make the question easier.
When we use this formula $\cot x=\dfrac{\cos x}{\sin x}$ to convert $\cot x$ in terms of $\sin x$ and $\cos x$.
We use this formula in the denominator of the question and convert into $\cos e{{c}^{2}}x$ and then convert in terms of $\sin x$.
$\cos e{{c}^{2}}x=1+{{\cot }^{2}}x$
We have a formula to convert $\cos ecx$ into $\sin x$
$\cos ecx=\dfrac{1}{\sin x}$
After putting the value of $\cot x=\dfrac{\cos x}{\sin x}$ and $\cos ecx=\dfrac{1}{\sin x}$ then, it can be written as:
$\dfrac{{{\cot }^{2}}x-1}{{{\cot }^{2}}x+1}=\dfrac{\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x}-1}{\cos e{{c}^{2}}x}=\dfrac{\dfrac{{{\cos }^{2}}x}{{{\sin }^{2}}x}-1}{\dfrac{1}{{{\sin }^{2}}x}}$
After taking the L.C.M in numerator and denominator, we get
$\dfrac{\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{\sin }^{2}}x}}{\dfrac{1}{{{\sin }^{2}}x}}$
Using this property
$\dfrac{\dfrac{a}{b}}{\dfrac{c}{d}}=\dfrac{a}{b}\times \dfrac{d}{c}$
It can be written as:
$\dfrac{{{\cos }^{2}}x-{{\sin }^{2}}x}{{{\sin }^{2}}x}\times {{\sin }^{2}}x$
After dividing by ${{\sin }^{2}}x$ we get
       ${{\cos }^{2}}x-{{\sin }^{2}}x$
We have to using this formula
${{\cos }^{2}}x-{{\sin }^{2}}x=\cos (x)\cos (x)-\sin (x)\sin (x)$
Then we use formula of trigonometric
$\cos A\cos B-\sin A\sin B=\cos (A+B)$
After using this formula, we get
${{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x$
Hence, we get
$\dfrac{{{\cot }^{2}}x-1}{{{\cot }^{2}}x+1}=\cos 2x$
Then $\dfrac{d}{dx}(\dfrac{{{\cot }^{2}}x-1}{{{\cot }^{2}}x+1})$ written as $\dfrac{d}{dx}(\cos 2x)$
We have to use the formula of differentiation of $\cos x$ with respect to x
$\dfrac{d}{dx}\left( \cos ax \right)=-a\sin ax$
After using the above formula, we get
$\dfrac{d}{dx}\left( \cos 2x \right)=-2\sin 2x$

Hence, $\dfrac{d}{dx}(\dfrac{{{\cot }^{2}}x-1}{{{\cot }^{2}}x+1})=-2\sin 2x$.

Note:
We should keep in mind all the trigonometric formulas and identities to solve any trigonometric function.
We use this trigonometric formula in above solutions are:
$\cot x=\dfrac{\cos x}{\sin x}$, $\cos e{{c}^{2}}x=1+{{\cot }^{2}}x$, $\cos ecx=\dfrac{1}{\sin x}$, ${{\cos }^{2}}x-{{\sin }^{2}}x=\cos 2x$
We should keep in mind all the formulas of differentiation and use carefully to solve the answer.
We use this differentiation formula in above solution are:
$\dfrac{d}{dx}\left( \cos ax \right)=-a\sin ax$.