Question

How do you differentiate \[y = \sqrt {\dfrac{{x - 1}}{{x + 1}}} \]

Answer 1

Hint: In this the function y is given we have to find the derivative of the function. The differentiation of a function is defined as the derivative or rate of change of a function. The function is said to be differentiable if the limit exists. Here in this question, we have to find derivatives with respect to x.

Complete step by step solution:
Consider the given function
 \[y = \sqrt {\dfrac{{x - 1}}{{x + 1}}} \]
This can be written as
 \[ \Rightarrow y = {\left( {\dfrac{{x - 1}}{{x + 1}}} \right)^{\dfrac{1}{2}}}\]
Applying the log on the both sides
 \[ \Rightarrow \log y = \log {\left( {\dfrac{{x - 1}}{{x + 1}}} \right)^{\dfrac{1}{2}}}\]
By the property of log \[\log {m^n} = n\log m\] , using this property the equation is written as
 \[ \Rightarrow \log y = \dfrac{1}{2}\log \left( {\dfrac{{x - 1}}{{x + 1}}} \right)\]
By the property of log \[\log \dfrac{m}{n} = \log m - \log n\] , using this property the equation is written as
 \[ \Rightarrow \log y = \dfrac{1}{2}\left( {\log (x - 1) - \log (x + 1)} \right)\]
On differentiating the above function with respect to x we have
 \[ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left[ {\dfrac{1}{{x - 1}} - \dfrac{1}{{x + 1}}} \right] \]
On simplifying we have
 \[ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left[ {\dfrac{{x + 1 - x + 1}}{{{x^2} - 1}}} \right] \]
On further simplifying the numerator we get
 \[ \Rightarrow \dfrac{1}{y}\dfrac{{dy}}{{dx}} = \dfrac{1}{2}\left[ {\dfrac{2}{{{x^2} - 1}}} \right] \]
On further simplification the equation is written as
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = y\left[ {\dfrac{1}{{{x^2} - 1}}} \right] \]
On substituting the value of y to the above equation
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \sqrt {\dfrac{{x - 1}}{{x + 1}}} \left[ {\dfrac{1}{{{x^2} - 1}}} \right] \]
In first term take root to both the numerator and denominator
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\sqrt {x - 1} }}{{\sqrt {x + 1} }}\dfrac{1}{{(x + 1)(x - 1)}}\]
On cancelling \[\sqrt {x - 1} \] we get
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {x + 1} \sqrt {x - 1} (x + 1)}}\]
We have standard algebraic formula \[(a + b)(a - b) = {a^2} - {b^2}\] , the above inequality is written as
 \[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {{x^2} - 1} (x + 1)}}\]
Therefore, we have \[\dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {{x^2} - 1} (x + 1)}}\]
Hence, we obtained the derivative.
So, the correct answer is “ \[\dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {{x^2} - 1} (x + 1)}}\] ”.

Note: The differentiation is the rate of change of a function at a point. We must know about the chain rule of derivatives. The function can be written as a composite of two functions, if the function can be written as a composite of two functions then we can apply the chain rule of derivative. By using log to the terms we can differentiate the function in an easy manner.