Question

If $\sec \left( {\dfrac{{x - y}}{{x + y}}} \right) = a$, then the value of $\dfrac{{dy}}{{dx}}$ is equal to:
(A) $\dfrac{y}{x}$
(B) $- \dfrac{y}{x}$
(C) $\dfrac{x}{y}$
(D) $- \dfrac{x}{y}$

Answer 1

Hint : In the given problem, we are required to differentiate $\sec \left( {\dfrac{{x - y}}{{x + y}}} \right) = a$ with respect to x. We will first simplify the function and then differentiate the function using the implicit method of differentiation. We must remember the power rule and chain rule of differentiation to get to the final answer.

Complete step-by-step answer :
So, we are given that $\sec \left( {\dfrac{{x - y}}{{x + y}}} \right) = a$.
So, we first take a secant inverse function on both sides of the equation. So, we get,
$\Rightarrow {\sec ^{ - 1}}\left[ {\sec \left( {\dfrac{{x - y}}{{x + y}}} \right)} \right] = {\sec ^{ - 1}}a$
Now, we know the formula ${\sec ^{ - 1}}\left( {\sec \left( x \right)} \right) = x$. So, we get,
$\Rightarrow \left( {\dfrac{{x - y}}{{x + y}}} \right) = {\sec ^{ - 1}}a$
Differentiating both sides of the equation with respect to x, we get,
$\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{x - y}}{{x + y}}} \right) = \dfrac{d}{{dx}}\left( {{{\sec }^{ - 1}}a} \right)$
Now, we can clearly see that the right side of the equation consists of only the constant terms. Also, we know that the derivative of a constant term with respect to x is zero. Hence, we get,
$\Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{{x - y}}{{x + y}}} \right) = 0$
Using the quotient rule of differentiation $\dfrac{d}{{dx}}\left( {\dfrac{{f(x)}}{{g(x)}}} \right) = \dfrac{{g(x) \times \dfrac{d}{{dx}}\left( {f(x)} \right) - f(x) \times \dfrac{d}{{dx}}\left( {g(x)} \right)}}{{{{\left[ {g\left( x \right)} \right]}^2}}}$, we get,
$\Rightarrow \dfrac{{\left( {x + y} \right)\dfrac{d}{{dx}}\left( {x - y} \right) - \left( {x - y} \right)\dfrac{d}{{dx}}\left( {x + y} \right)}}{{{{\left( {x + y} \right)}^2}}} = 0$
We use the power rule of differentiation, we get,
$\Rightarrow \dfrac{{\left( {x + y} \right)\left( {1 - \dfrac{{dy}}{{dx}}} \right) - \left( {x - y} \right)\left( {1 + \dfrac{{dy}}{{dx}}} \right)}}{{{{\left( {x + y} \right)}^2}}} = 0$
Cross multiplying the terms of the equation, we get,
$\Rightarrow \left( {x + y} \right)\left( {1 - \dfrac{{dy}}{{dx}}} \right) - \left( {x - y} \right)\left( {1 + \dfrac{{dy}}{{dx}}} \right) = 0$
Opening the brackets, we get,
$\Rightarrow x + y - x\dfrac{{dy}}{{dx}} - y\dfrac{{dy}}{{dx}} - x + y - x\dfrac{{dy}}{{dx}} + y\dfrac{{dy}}{{dx}} = 0$
Cancelling like terms with opposite signs, we get,
$\Rightarrow y - x\dfrac{{dy}}{{dx}} + y - x\dfrac{{dy}}{{dx}} = 0$
Simplifying the equation further, we get,
$\Rightarrow 2x\dfrac{{dy}}{{dx}} = 2y$
Finding the value of $\dfrac{{dy}}{{dx}}$, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{y}{x}$
Hence, option (A) is the correct answer.
So, the correct answer is “Option A”.

Note : Implicit functions are those functions that involve two variables and the two variables are not separable and cannot be isolated from each other. Hence, we have to follow a certain method to differentiate such functions and also apply the chain rule of differentiation. We must remember the simple derivatives of basic functions to solve such problems.