Question

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

Answer 1

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

Complete step-by-step answer:
\[f(x)=\dfrac{\cot x}{1+\csc x}\]. . . . . . . . . . . . . . . . . . . . . (a)
\[\dfrac{d}{dx}\left( \cot x \right)=-{{\csc }^{2}}x\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
\[\dfrac{d}{dx}\left( \csc x \right)=-\csc x\cot x\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (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
\[=\dfrac{\left( 1+\csc x \right)\left( -{{\csc }^{2}}x \right)-\cot x\left( \dfrac{d}{dx}\left( \csc x \right) \right)}{{{\left( 1+\csc x \right)}^{2}}}\]
Further solving we get the derivative of the function as
\[=\dfrac{\left( 1+\csc x \right)\left( -{{\csc }^{2}}x \right)-\cot x\left( -\csc x\cot x \right)}{{{\left( 1+\csc x \right)}^{2}}}\]
Further solving we get the derivative of the function as
\[=\dfrac{-\left( 1+\csc x \right)\left( {{\csc }^{2}}x \right)+\csc x{{\cot }^{2}}x}{{{\left( 1+\csc x \right)}^{2}}}\]
Further solving we get the derivative of the function as
\[=\dfrac{-\left( 1+\csc x \right)\left( {{\csc }^{2}}x \right)+\csc x{{\cot }^{2}}x}{{{\left( 1+\csc x \right)}^{2}}}\]
Further solving we get the derivative of the function as
\[=\dfrac{-\left( {{\csc }^{2}}x \right)-\csc x\left( {{\csc }^{2}}x \right)+\csc x{{\cot }^{2}}x}{{{\left( 1+\csc x \right)}^{2}}}\]
\[=\dfrac{-\left( {{\csc }^{2}}x \right)-\csc x({{\csc }^{2}}x-{{\cot }^{2}}x)}{{{\left( 1+\csc x \right)}^{2}}}\]
\[=\dfrac{-\left( {{\csc }^{2}}x \right)-\csc x(1)}{{{\left( 1+\csc x \right)}^{2}}}\]
\[=-\dfrac{\left( {{\csc }^{2}}x \right)+\csc x}{{{\left( 1+\csc x \right)}^{2}}}\]

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 \[\dfrac{d}{dx}\left( \dfrac{f}{g} \right)=\dfrac{{{f}^{1}}g-f{{g}^{1}}}{{{g}^{2}}}\]. If we are doing derivative means we are finding the slope of a function. Care should be taken while doing calculations.