Question

How do you differentiate $f(x) = \dfrac{1}{{\cot (x)}}$using the chain rule?

Answer 1

Hint: In the question, we will be using the chain rule which is given as,
\[\dfrac{d}{{dx}}[f(g(x))] = f'\{ g(x)\} .g'(x)\]. DIrectly solving by taking tanx and finding derivatives is not desired method here.

Complete step-by-step answer:
Here, $g(x) = \cot x$and$f(x) = \dfrac{1}{x}$,
So, according to the chain rule we have two parts to solve, one is \[f'\{ g(x)\} \]and other is \[g'(x)\]
Let us first find the value of\[g'(x)\],
As we have taken \[g(x) = \cot x\]and we know that$\cot x = \dfrac{{\cos x}}{{\sin x}}$, so \[g'(x)\]will be calculated by,
\[\dfrac{d}{{dx}}(\dfrac{u}{v}) = \dfrac{{u'v - uv'}}{{{v^2}}}\]
Here u is \[cosx\]and v is \[sinx\]so $u' = - \sin x$and$v' = \cos x$, putting these in above equation w3e will get,
\[\dfrac{d}{{dx}}(\dfrac{u}{v}) = \dfrac{{ - \sin x \times \sin x - \cos x \times \cos x}}{{{{\sin }^2}x}}\]
which will be equal to,
\[\dfrac{d}{{dx}}(\dfrac{u}{v}) = \dfrac{{ - {{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x}}\]
Taking minus sign common in the numerator we will have,
\[\dfrac{d}{{dx}}(\dfrac{u}{v}) = \dfrac{{ - ({{\sin }^2}x + {{\cos }^2}x)}}{{{{\sin }^2}x}}\],
As we all know that\[{\sin ^2}x + {\cos ^2}x = 1\], than the above equation will become,
\[\dfrac{d}{{dx}}(\dfrac{u}{v}) = \dfrac{{ - 1}}{{{{\sin }^2}x}}\]
And also, \[\dfrac{1}{{{{\sin }^2}x}} = \cos e{c^2}x\]
\[\dfrac{d}{{dx}}(\dfrac{u}{v}) = - \cos e{c^2}x\]
Now let us find f’(x),
As\[f\left( x \right) = \dfrac{1}{x}\], we can also write it as $f(x) = {x^{ - 1}}$
And by using the identity, \[\dfrac{d}{{dx}}({x^n}) = n{x^{n - 1}}\], we will get
\[\dfrac{d}{{dx}}({x^{ - 1}}) = ( - 1){x^{ - 1 - 1}}\]
\[\dfrac{d}{{dx}}({x^{ - 1}}) = ( - 1){x^{ - 2}}\]
Now \[f'\{ g(x)\} \]= $( - 1){(\cot x)^{ - 2}}$than
\[f'\{ g(x)\} .g'(x)\]=$( - 1){(\cot x)^{ - 2}} \times ( - \cos e{c^2}x)$
=$\dfrac{{coe{{\sec }^2}x}}{{{{\cot }^2}x}}$
So, the correct answer is “$\dfrac{{coe{{\sec }^2}x}}{{{{\cot }^2}x}}$”.

Note: In mathematics, the derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus.