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Question: Differentiate \[\dfrac{d}{dx}\left( \dfrac{1+\cot x}{1-\cot x} \right)\] ....

Differentiate ddx(1+cot⁡x1−cot⁡x)\dfrac{d}{dx}\left( \dfrac{1+\cot x}{1-\cot x} \right) .

Explanation

Solution

In this we have to differentiate the given equation. We can first differentiate the given expression in uv\dfrac{u}{v} method, which is uv=vu′−uv′v2\dfrac{u}{v}=\dfrac{vu'-uv'}{{{v}^{2}}}. We can then simplify the terms by taking the common terms and cancelling it. We can then use some trigonometric formulas and identities to simplify the equation step by step to get the simplified differentiated solution.

Complete step by step answer:
Here we have to differentiate the given expression,
ddx(1+cot⁡x1−cot⁡x)\dfrac{d}{dx}\left( \dfrac{1+\cot x}{1-\cot x} \right)
We can now differentiate the above expression using uv\dfrac{u}{v} method.
We know that the uv\dfrac{u}{v} method formula is,
uv=vu′−uv′v2\dfrac{u}{v}=\dfrac{vu'-uv'}{{{v}^{2}}}
We can now differentiate the given expression, we get
⇒(1−cot⁡x)(1+cot⁡x)′−(1+cot⁡x)(1−cot⁡x)′(1−cot⁡x)2\Rightarrow \dfrac{\left( 1-\cot x \right)\left( 1+\cot x \right)'-\left( 1+\cot x \right)\left( 1-\cot x \right)'}{{{\left( 1-\cot x \right)}^{2}}}
We can now differentiate the numerator, we get
⇒(1−cot⁡x)(0−csc⁡2x)−(1+cot⁡x)(0+csc⁡2x)(1−cot⁡x)2\Rightarrow \dfrac{\left( 1-\cot x \right)\left( 0-{{\csc }^{2}}x \right)-\left( 1+\cot x \right)\left( 0+{{\csc }^{2}}x \right)}{{{\left( 1-\cot x \right)}^{2}}}
We can now simplify the above terms, we get
⇒(1−cot⁡x)(−csc⁡2x)−(1+cot⁡x)(csc⁡2x)(1−cot⁡x)2\Rightarrow \dfrac{\left( 1-\cot x \right)\left( -{{\csc }^{2}}x \right)-\left( 1+\cot x \right)\left( {{\csc }^{2}}x \right)}{{{\left( 1-\cot x \right)}^{2}}}
We can now take the common term in the numerator, we get
⇒(−csc⁡2x)(1+cot⁡x+1−cot⁡x)(1−cot⁡x)2=−2csc⁡2x(1−cot⁡x)2\Rightarrow \dfrac{\left( -{{\csc }^{2}}x \right)\left( 1+\cot x+1-\cot x \right)}{{{\left( 1-\cot x \right)}^{2}}}=\dfrac{-2{{\csc }^{2}}x}{{{\left( 1-\cot x \right)}^{2}}}
We can now split the denominator using the algebraic whole square formula, we get
⇒−2csc⁡2x1+cot⁡2x−2cot⁡x\Rightarrow \dfrac{-2{{\csc }^{2}}x}{1+{{\cot }^{2}}x-2\cot x}
We know that 1+cot⁡2x=csc⁡2x1+{{\cot }^{2}}x={{\csc }^{2}}x, we can substitute in the above step, we get
⇒−2csc⁡2xcsc⁡2x−2cot⁡x\Rightarrow \dfrac{-2{{\csc }^{2}}x}{{{\csc }^{2}}x-2\cot x}
We know that, we can apply this in the above step, we get
⇒−2sin⁡2x1sin⁡2x−2cos⁡xsin⁡x\Rightarrow \dfrac{\dfrac{-2}{{{\sin }^{2}}x}}{\dfrac{1}{{{\sin }^{2}}x}-2\dfrac{\cos x}{\sin x}}
We can now simplify the above step, we get
⇒−2sin⁡2xsin⁡x−2cos⁡xsin⁡2xsin⁡3x=−2sin⁡2x1−2cos⁡xsin⁡xsin⁡2x\Rightarrow \dfrac{\dfrac{-2}{{{\sin }^{2}}x}}{\dfrac{\sin x-2\cos x{{\sin }^{2}}x}{{{\sin }^{3}}x}}=\dfrac{\dfrac{-2}{{{\sin }^{2}}x}}{\dfrac{1-2\cos x\sin x}{{{\sin }^{2}}x}}
We can now cancel the similar terms in both the denominator, we get
⇒−21−2cos⁡xsin⁡x=−21−sin⁡2x ∵2cos⁡xsin⁡x=sin⁡2x\Rightarrow \dfrac{-2}{1-2\cos x\sin x}=\dfrac{-2}{1-\sin 2x}\text{ }\because 2\cos x\sin x=\sin 2x
Therefore, ddx(1+cot⁡x1−cot⁡x)=−2sin⁡2x−1\dfrac{d}{dx}\left( \dfrac{1+\cot x}{1-\cot x} \right)=\dfrac{-2}{\sin 2x-1}.

Note: We should always remember that when we are given a fraction of terms to be differentiated then we can use the uv\dfrac{u}{v} method, uv=vu′−uv′v2\dfrac{u}{v}=\dfrac{vu'-uv'}{{{v}^{2}}} to differentiate it. We should also remember that the trigonometric formulas like csc⁡2x=1sin⁡2x,cot⁡x=cos⁡xsin⁡x{{\csc }^{2}}x=\dfrac{1}{{{\sin }^{2}}x},\cot x=\dfrac{\cos x}{\sin x} and 1+cot⁡2x=csc⁡2x1+{{\cot }^{2}}x={{\csc }^{2}}x to simplify the given differentiation and to get the final answer.