Solveeit Logo
Login

Question

Question: Let us define a function as, \(f\left( x \right) = \sqrt {ax} + \dfrac{{{a^2}}}{{\sqrt {ax} }}\). No...

Let us define a function as, f(x)=ax+a2axf\left( x \right) = \sqrt {ax} + \dfrac{{{a^2}}}{{\sqrt {ax} }}. Now choose the value of f1(a){f^1}\left( a \right) from the below options.

  1. −1 - 1
  2. 00
  3. 11
  4. aa
Explanation

Solution

Hint : Given a function to find its differentiation. To differentiate a function we first need to rewrite the given function into a suitable form so that it is easy to differentiate. We can differentiate a function in two ways: that is by direct definition method and the other is by the chain rule. It is easy to use the chain rule method. But to use the chain rule we must need to know some differentials for basic functions. After that, we can easily differentiate using the chain rule.

Complete step-by-step answer :
Given that,
f(x)=ax+a2axf\left( x \right) = \sqrt {ax} + \dfrac{{{a^2}}}{{\sqrt {ax} }}.
Now let us rewrite it in the suitable form
f(x)=ax12+a2ax−12f\left( x \right) = \sqrt a {x^{\dfrac{1}{2}}} + \dfrac{{{a^2}}}{{\sqrt a }}{x^{ - \dfrac{1}{2}}}
Now let to differentiate this function we need to know some differentials,
So, we know the power rule of differentiation ddx(xn)=n.xn−1\dfrac{d}{{dx}}\left( {{x^n}} \right) = n.{x^{n - 1}}. So, we get,
Now let us use this formula to differentiate the given function.
By chain rule we get,
⇒f′(x)=a×ddx(x12)+a2a×ddx(x−12)\Rightarrow f'\left( x \right) = \sqrt a \times \dfrac{d}{{dx}}\left( {{x^{\dfrac{1}{2}}}} \right) + \dfrac{{{a^2}}}{{\sqrt a }} \times \dfrac{d}{{dx}}\left( {{x^{ - \dfrac{1}{2}}}} \right) ,
After using the formula, we get
⇒f′(x)=a×12(x12−1)+a2a×−12×(x−12−1)\Rightarrow f'\left( x \right) = \sqrt a \times \dfrac{1}{2}\left( {{x^{\dfrac{1}{2} - 1}}} \right) + \dfrac{{{a^2}}}{{\sqrt a }} \times - \dfrac{1}{2} \times \left( {{x^{ - \dfrac{1}{2} - 1}}} \right)
On simplification we get,
⇒f′(x)=a2(x−12)−aa2(x−32)\Rightarrow f'\left( x \right) = \dfrac{{\sqrt a }}{2}\left( {{x^{ - \dfrac{1}{2}}}} \right) - \dfrac{{a\sqrt a }}{2}\left( {{x^{ - \dfrac{3}{2}}}} \right)
Now, let us substitute the value aa in the function f′(x)f'\left( x \right) .
We get,
⇒f′(x)=a2(a−12)−aa2(a−32)\Rightarrow f'\left( x \right) = \dfrac{{\sqrt a }}{2}\left( {{a^{ - \dfrac{1}{2}}}} \right) - \dfrac{{a\sqrt a }}{2}\left( {{a^{ - \dfrac{3}{2}}}} \right),
Cancelling the common factor in numerator and denominator, we get
⇒f′(x)=12−12\Rightarrow f'\left( x \right) = \dfrac{1}{2} - \dfrac{1}{2}
⇒f′(x)=0\Rightarrow f'\left( x \right) = 0
Observe that it does not depend on the value of aa .
It means f′(x)f'\left( x \right) is independent of aa .
So, the correct answer is “Option B”.

Note : This is a good practice example for using the chain rule. We cannot use the definition of the differential all the time to find the differentiation of a function because of the time constraint. We need to reduce the time in the exam. By answer, we can also say that at the point (a,2a)\left( {a,2a} \right) on the graph has local minima or local maxima since the value of f1(a){f^1}\left( a \right) is 00.