Question

If we have an expression as $xy = {tan^{-1}}(x \times y) + {cot^{-1}}(x \times y),$ then \[\dfrac{{dy}}{{dx}}\]is equal to
\[1)\]$\dfrac{y}{x}$
\[2)\]$\dfrac{{ - y}}{x}$
\[3)\]\[\dfrac{x}{y}\]
\[4)\]\[\dfrac{{ - x}}{y}\]

Answer 1

Hint: We have to find the derivative of [$xy = {tan^{-1}}(x \times y) + {cot^{-1}}(x \times y),$ ] with respect to x. We solve this question using the product rule of differentiation and using various basic derivative formulas of trigonometric functions and derivatives of${x^n}$. We firstly use the property of the inverse of the trigonometric function to get a relation between $x$ and $y$ . Then by differentiating the expression we get the value of\[\dfrac{{dy}}{{dx}}\].

Complete step-by-step solution:
Differentiation, in mathematics , is the process of finding the derivative , or the rate of change of a given function . In contrast to the abstract nature of the theory behind it , the practical technique of differentiation can be carried out by purely algebraic manipulations , using three basic derivatives , four rules of operation , and a knowledge of how to manipulate functions. We can solve any of the problems using the rules of operations i.e. addition , subtraction , multiplication and division .
Given : $xy = {tan^{-1}}(x \times y) + {cot^{-1}}(x \times y),$
We know that , properties of inverse trigonometry :
${tan^{-1}}(x) + {cot^{-1}}(x) = \dfrac{\pi }{2},x \in R$
Where $R$ is the set of real numbers
Using this property of inverse trigonometry , we get
${tan^{-1}}(x \times y) + {cot^{-1}}(x \times y) = \dfrac{\pi }{2}$
Now , the expression becomes
\[x{\text{ }} \times {\text{ }}y{\text{ }} = \;\dfrac{\pi }{2}\]
Now we have the derivative of $y$ with respect to x.
Derivative of product of two function is given by the following product rule :
\[{\text{ }}\dfrac{{d\left[ {f\left( x \right){\text{ }} \times {\text{ }}g\left( x \right)} \right]}}{{dx}} = {\text{ }}\dfrac{{d\left[ {f\left( x \right)} \right]}}{{dx}}{\text{ }} \times {\text{ }}g{\text{ }} + {\text{ }}f{\text{ }} \times {\text{ }}\dfrac{{d\left[ {g\left( x \right)} \right]}}{{dx}}\]
 ( derivative of${x^n} = n \times {x^{(n - 1)}}$)
( derivative of constant\[ = {\text{ }}0\])
Using the product rule , we get
\[x{\text{ }} \times {\text{ }}\dfrac{{dy}}{{dx}} + {\text{ }}y{\text{ }} = {\text{ }}0\]
Simplifying the term , we get
\[\dfrac{{dy}}{{dx}}{\text{ }} = {\text{ }}\dfrac{{ - y}}{x}\]
Hence , the derivative of$xy = {tan^{-1}}(x \times y) + {cot^{-1}}(x \times y),$ with respect to x is \[\dfrac{{ - y}}{x}\]
Thus , the correct option is \[\left( 2 \right)\]

Note: We differentiated y with respect to to find \[\dfrac{{dy}}{{dx}}\] . We know the differentiation of trigonometric function :
\[\dfrac{{d\left[ {cos{\text{ }}x} \right]}}{{dx}} = {\text{ }} - sin{\text{ }}x\]
\[\dfrac{{d\left[ {sin{\text{ }}x} \right]{\text{ }}}}{{dx}}{\text{ }} = {\text{ }}cos{\text{ }}x\]
$d[{x^n}] = n{x^{(n - 1)}}$
$d[\tan x] = {sec^{2}}x$
We use the derivative according to the given problem .