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Question: Find \[{{f}^{1}}\left( x \right)\]. \[f(x)=\dfrac{\cot x}{1+\csc x}\]...

Find f1(x){{f}^{1}}\left( x \right).
f(x)=cot⁡x1+csc⁡xf(x)=\dfrac{\cot x}{1+\csc x}

Explanation

Solution

Hint: To solve the above problem we have to know the basic derivatives of cot⁡x\cot xand csc⁡x\csc x. After writing the derivatives rewrite the equation with the derivatives of the function.
ddx(cot⁡x)=−csc⁡2x\dfrac{d}{dx}\left( \cot x \right)=-{{\csc }^{2}}x, ddx(csc⁡x)=−csc⁡xcot⁡x\dfrac{d}{dx}\left( \csc x \right)=-\csc x\cot x. We can see one function is inside another we have to find internal derivatives.

Complete step-by-step answer:
f(x)=cot⁡x1+csc⁡xf(x)=\dfrac{\cot x}{1+\csc x}. . . . . . . . . . . . . . . . . . . . . (a)
ddx(cot⁡x)=−csc⁡2x\dfrac{d}{dx}\left( \cot x \right)=-{{\csc }^{2}}x. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
ddx(csc⁡x)=−csc⁡xcot⁡x\dfrac{d}{dx}\left( \csc x \right)=-\csc x\cot x. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Substituting (1) and (2) as a derivative we get,
Therefore derivative of the given function is, this is solved by quotient rule,
{{f}^{1}}(x)$$$$=\dfrac{\left( 1+\csc x \right)\dfrac{d}{dx}\left( \cot x \right)-\cot x\left( \dfrac{d}{dx}\left( 1+\csc x \right) \right)}{{{\left( 1+\csc x \right)}^{2}}}
Further solving we get the derivative of the function as
=(1+csc⁡x)(−csc⁡2x)−cot⁡x(ddx(csc⁡x))(1+csc⁡x)2=\dfrac{\left( 1+\csc x \right)\left( -{{\csc }^{2}}x \right)-\cot x\left( \dfrac{d}{dx}\left( \csc x \right) \right)}{{{\left( 1+\csc x \right)}^{2}}}
Further solving we get the derivative of the function as
=(1+csc⁡x)(−csc⁡2x)−cot⁡x(−csc⁡xcot⁡x)(1+csc⁡x)2=\dfrac{\left( 1+\csc x \right)\left( -{{\csc }^{2}}x \right)-\cot x\left( -\csc x\cot x \right)}{{{\left( 1+\csc x \right)}^{2}}}
Further solving we get the derivative of the function as
=−(1+csc⁡x)(csc⁡2x)+csc⁡xcot⁡2x(1+csc⁡x)2=\dfrac{-\left( 1+\csc x \right)\left( {{\csc }^{2}}x \right)+\csc x{{\cot }^{2}}x}{{{\left( 1+\csc x \right)}^{2}}}
Further solving we get the derivative of the function as
=−(1+csc⁡x)(csc⁡2x)+csc⁡xcot⁡2x(1+csc⁡x)2=\dfrac{-\left( 1+\csc x \right)\left( {{\csc }^{2}}x \right)+\csc x{{\cot }^{2}}x}{{{\left( 1+\csc x \right)}^{2}}}
Further solving we get the derivative of the function as
=−(csc⁡2x)−csc⁡x(csc⁡2x)+csc⁡xcot⁡2x(1+csc⁡x)2=\dfrac{-\left( {{\csc }^{2}}x \right)-\csc x\left( {{\csc }^{2}}x \right)+\csc x{{\cot }^{2}}x}{{{\left( 1+\csc x \right)}^{2}}}
=−(csc⁡2x)−csc⁡x(csc⁡2x−cot⁡2x)(1+csc⁡x)2=\dfrac{-\left( {{\csc }^{2}}x \right)-\csc x({{\csc }^{2}}x-{{\cot }^{2}}x)}{{{\left( 1+\csc x \right)}^{2}}}
=−(csc⁡2x)−csc⁡x(1)(1+csc⁡x)2=\dfrac{-\left( {{\csc }^{2}}x \right)-\csc x(1)}{{{\left( 1+\csc x \right)}^{2}}}
=−(csc⁡2x)+csc⁡x(1+csc⁡x)2=-\dfrac{\left( {{\csc }^{2}}x \right)+\csc x}{{{\left( 1+\csc x \right)}^{2}}}

Note: In the above problem we have solved using the Quotient rule. If both the functions f and g are differentiable then Quotient rule is given by ddx(fg)=f1g−fg1g2\dfrac{d}{dx}\left( \dfrac{f}{g} \right)=\dfrac{{{f}^{1}}g-f{{g}^{1}}}{{{g}^{2}}}. If we are doing derivative means we are finding the slope of a function. Care should be taken while doing calculations.