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Question: If \(y = \dfrac{{[\sqrt {(a + x)} - \sqrt {(a - x)} ]}}{{[\sqrt {(a + x)} + \sqrt {(a - x)} ]}}\) th...

If y=[(a+x)−(a−x)][(a+x)+(a−x)]y = \dfrac{{[\sqrt {(a + x)} - \sqrt {(a - x)} ]}}{{[\sqrt {(a + x)} + \sqrt {(a - x)} ]}} then dydx=\dfrac{{dy}}{{dx}} = ?
A) ayx(a2−x2)\dfrac{{ay}}{{x\sqrt {({a^2} - {x^2})} }}
B) ay(a2−x2)\dfrac{{ay}}{{\sqrt {({a^2} - {x^2})} }}
C) ayx(x2−a2)\dfrac{{ay}}{{x\sqrt {({x^2} - {a^2})} }}
D) None of these

Explanation

Solution

In this question given an expression and we have to find its derivative. To find the solution for the problem we first multiply the numerator and denominator by (a+x)−(a−x)\sqrt {(a + x)} - \sqrt {(a - x)} . Then we will simplify the equation and after that we will differentiate the equation with respect to xx and by doing some more mathematical steps we will reach our final answer.

Complete step by step answer:
First we will see the given equation,
⇒y=[(a+x)−(a−x)][(a+x)+(a−x)]\Rightarrow y = \dfrac{{[\sqrt {(a + x)} - \sqrt {(a - x)} ]}}{{[\sqrt {(a + x)} + \sqrt {(a - x)} ]}}
Now, we will do multiplication in numerator and denominator by (a+x)−(a−x)\sqrt {(a + x)} - \sqrt {(a - x)} . In short, we will take conjugate.
⇒[(a+x)−(a−x)][(a+x)+(a−x)]×(a+x)−(a−x)(a+x)−(a−x)\Rightarrow \dfrac{{[\sqrt {(a + x)} - \sqrt {(a - x)} ]}}{{[\sqrt {(a + x)} + \sqrt {(a - x)} ]}} \times \dfrac{{\sqrt {(a + x)} - \sqrt {(a - x)} }}{{\sqrt {(a + x)} - \sqrt {(a - x)} }}
Now, we need to do some simplification in above expression,
⇒[(a+x)−(a−x)]2[a+x−a+x]\Rightarrow \dfrac{{{{[\sqrt {(a + x)} - \sqrt {(a - x)} ]}^2}}}{{[a + x - a + x]}}
⇒[(a+x)−(a−x)]22x\Rightarrow \dfrac{{{{[\sqrt {(a + x)} - \sqrt {(a - x)} ]}^2}}}{{2x}}
Now, open bracket of numerator,
⇒[a+x+a−x−2(a+x)(a−x)2x\Rightarrow \dfrac{{[a + x + a - x - 2\sqrt {(a + x)} \sqrt {(a - x)} }}{{2x}}
⇒[2a−2a2−x2]2x\Rightarrow \dfrac{{[2a - 2\sqrt {{a^2} - {x^2}} ]}}{{2x}}
Now, taking common 2 from numerator and cancelling it with denominator and we will get,
⇒[a−a2−x2]x\Rightarrow \dfrac{{[a - \sqrt {{a^2} - {x^2}} ]}}{x}
So we get,
⇒y=[a−a2−x2]x\Rightarrow y = \dfrac{{[a - \sqrt {{a^2} - {x^2}} ]}}{x}
Now call the above equation as equation number ii .
Now, differentiate equation ii with respect to xx
dydx=[x(−12a2−x2x−2x−a(a−a2−x2))]x2\dfrac{{dy}}{{dx}} = \dfrac{{[x( - \dfrac{1}{2}\sqrt {{a^2} - {x^2}} x - 2x - a(a - \sqrt {{a^2} - {x^2}} ))]}}{{{x^2}}}
We need to do some simplification on above equation,
dydx=[x2a2−x2−a+a2−x2]x2\dfrac{{dy}}{{dx}} = \dfrac{{[{x^2}\sqrt {{a^2} - {x^2}} - a + \sqrt {{a^2} - {x^2}} ]}}{{{x^2}}}
dydx=[a2−aa2−x2]x2a2−x2\dfrac{{dy}}{{dx}} = \dfrac{{[{a^2} - a\sqrt {{a^2} - {x^2}} ]}}{{{x^2}\sqrt {{a^2} - {x^2}} }}
Take aa common from numerator and we will get,
dydx=a[a−a2−x2]x2a2−x2\dfrac{{dy}}{{dx}} = \dfrac{{a[a - \sqrt {{a^2} - {x^2}} ]}}{{{x^2}\sqrt {{a^2} - {x^2}} }}
Compare above equation with equation ii
⇒dydx=ayxa2−x2\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ay}}{{x\sqrt {{a^2} - {x^2}} }}
Therefore, the correct answer is option (A) ayx(a2−x2)\dfrac{{ay}}{{x\sqrt {({a^2} - {x^2})} }}.

Note:
In such types of questions, students might make mistakes to calculate simple mathematics simplifications then they find that they will not get the right answer. So, in these types of questions students have to do all simple steps to bypass the simple errors so they will get the right answer.