Solveeit Logo
Login

Question

Question: If we have an expression as \(xy = {tan^{-1}}(x \times y) + {cot^{-1}}(x \times y),\) then \[\dfrac{...

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

Explanation

Solution

We have to find the derivative of [xy=tan−1(x×y)+cot−1(x×y),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 ofxn{x^n}. We firstly use the property of the inverse of the trigonometric function to get a relation between xx and yy . Then by differentiating the expression we get the value ofdydx\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×y)+cot−1(x×y),xy = {tan^{-1}}(x \times y) + {cot^{-1}}(x \times y),
We know that , properties of inverse trigonometry :
tan−1(x)+cot−1(x)=π2,x∈R{tan^{-1}}(x) + {cot^{-1}}(x) = \dfrac{\pi }{2},x \in R
Where RR is the set of real numbers
Using this property of inverse trigonometry , we get
tan−1(x×y)+cot−1(x×y)=π2{tan^{-1}}(x \times y) + {cot^{-1}}(x \times y) = \dfrac{\pi }{2}
Now , the expression becomes
x × y =  π2x{\text{ }} \times {\text{ }}y{\text{ }} = \;\dfrac{\pi }{2}
Now we have the derivative of yy with respect to x.
Derivative of product of two function is given by the following product rule :
 d[f(x) × g(x)]dx= d[f(x)]dx × g + f × d[g(x)]dx{\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 ofxn=n×x(n−1){x^n} = n \times {x^{(n - 1)}})
( derivative of constant= 0 = {\text{ }}0)
Using the product rule , we get
x × dydx+ y = 0x{\text{ }} \times {\text{ }}\dfrac{{dy}}{{dx}} + {\text{ }}y{\text{ }} = {\text{ }}0
Simplifying the term , we get
dydx = −yx\dfrac{{dy}}{{dx}}{\text{ }} = {\text{ }}\dfrac{{ - y}}{x}
Hence , the derivative ofxy=tan−1(x×y)+cot−1(x×y),xy = {tan^{-1}}(x \times y) + {cot^{-1}}(x \times y), with respect to x is −yx\dfrac{{ - y}}{x}
Thus , the correct option is (2)\left( 2 \right)

Note: We differentiated y with respect to to find dydx\dfrac{{dy}}{{dx}} . We know the differentiation of trigonometric function :
d[cos x]dx= −sin x\dfrac{{d\left[ {cos{\text{ }}x} \right]}}{{dx}} = {\text{ }} - sin{\text{ }}x
d[sin x] dx = cos x\dfrac{{d\left[ {sin{\text{ }}x} \right]{\text{ }}}}{{dx}}{\text{ }} = {\text{ }}cos{\text{ }}x
d[xn]=nx(n−1)d[{x^n}] = n{x^{(n - 1)}}
d[tan⁡x]=sec2xd[\tan x] = {sec^{2}}x
We use the derivative according to the given problem .