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.