Question

Simplify- $\dfrac{d}{{dx}}\left( {\dfrac{{{{\cot }^2}x - 1}}{{{{\cot }^2}x + 1}}} \right)$

Answer 1

In this question we can see that we have trigonometric expressions. So here we will apply the trigonometric identities or formulas to solve this question. We will first write the cotangent in the form of the ratio of sine function to cosine function i.e. $\cot x = \dfrac{{\sin x}}{{\cos x}}$ . By applying this we will simplify the given function. Then we will find the derivative of that simplified function.

Formula used:
${\sin ^2}x + {\cos ^2}x = 1$
$\cos 2x = {\cos ^2}x - {\sin ^2}x$ .

Complete step-by-step answer:
We have been given
$\dfrac{d}{{dx}}\left( {\dfrac{{{{\cot }^2}x - 1}}{{{{\cot }^2}x + 1}}} \right)$
Let us assume that
$\Rightarrow y = \left( {\dfrac{{{{\cot }^2}x - 1}}{{{{\cot }^2}x + 1}}} \right)$
By applying the basic formula of trigonometric ratio, we can write ${\cot ^2}x$ as $\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}}$
By putting this value in the expression, we can write it as:
$= \left( {\dfrac{{\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - 1}}{{\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - 1}}} \right)$
We will now take the LCM of the above and simplify this:
$= \left( {\dfrac{{\dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\cos }^2}x}}}}{{\dfrac{{{{\sin }^2}x + {{\cos }^2}x}}{{{{\cos }^2}x}}}}} \right)$
The common factor from the denominators will get cancelled, so we have:
$= \left( {\dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\sin }^2}x + {{\cos }^2}x}}} \right)$
Now we will substitute the values of these by the formula mentioned above, so we have:
$\Rightarrow y = \dfrac{{\cos 2x}}{1}$
We will now find the derivative of the function above, so we can write it as:
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\cos 2x} \right)$
Now we know that the derivative of the cosine function:
$\dfrac{{dy}}{{dx}}(\cos x) = - \sin x$ ,
Therefore the differentiation of
$\dfrac{d}{{dx}}(\cos 2x) = - \sin 2x$
And we have another term, so the differentiation of $2x$ is $2$
By putting these value together we have:
$\Rightarrow \dfrac{{dy}}{{dx}} = - \sin 2x \times 2$
It gives us value $\Rightarrow \dfrac{{dy}}{{dx}} = - 2 \cdot \sin 2x$

Note: We should note that to find the derivative of $2x$ , we have applied a very basic formula. We know that the derivative of the form $cx$ , where $c$ is a constant, is given by : $\left[ {cx} \right]' = c$ . Since the derivative of $cx$ is $c$ , so by applying this formula we can say that the derivative of the function $2x$ is $2$ . We should always remember the trigonometric formula to solve the question easily in a simpler way by avoiding calculation mistakes.