Question

If $x = \left( {\dfrac{{1 - \sqrt y }}{{1 + \sqrt y }}} \right),$ then $\dfrac{{dy}}{{dx}}$ is equal to ;
$\left( 1 \right)\dfrac{4}{{{{\left( {x + 1} \right)}^2}}}$
$\left( 2 \right)\dfrac{{4\left( {x - 1} \right)}}{{{{\left( {x + 1} \right)}^3}}}$
$\left( 3 \right)\dfrac{{\left( {x - 1} \right)}}{{{{\left( {x + 1} \right)}^3}}}$
$\left( 4 \right)\dfrac{4}{{{{\left( {x + 1} \right)}^3}}}$

Answer 1

Hint: This question requires the knowledge of basic derivative formulae and algebraic identities. We have to calculate $\dfrac{{dy}}{{dx}}$ which determines the rate of change of $y$ with respect to $x$ . Here the independent variable is $x$ and $y$ is the dependent variable. But we are given the value of $x$ in terms of $y$ i.e. $x = \left( {\dfrac{{1 - \sqrt y }}{{1 + \sqrt y }}} \right),$ so first we will get the expression of $y$ in terms of $x$ and then we will differentiate it with respect to $x$ .

Complete step by step solution:
The given expression is ;
$ \Rightarrow x = \left( {\dfrac{{1 - \sqrt y }}{{1 + \sqrt y }}} \right)$
But we have to calculate $\dfrac{{dy}}{{dx}}$ , so we need to get $y{\text{ in terms of }}x$ from the given expression;
By cross multiplying the given expression;
$ \Rightarrow x\left( {1 + \sqrt y } \right) = \left( {1 - \sqrt y } \right)$
$ \Rightarrow x + x\sqrt y = 1 - \sqrt y $
Taking the $\sqrt y $ terms to the L.H.S. ;
$ \Rightarrow x\sqrt y + \sqrt y = 1 - x$
Taking $\sqrt y $ common on L.H.S. ;
$ \Rightarrow \sqrt y \left( {x + 1} \right) = 1 - x$
Since we have to calculate the value of $y$ in terms of $x$ , therefore taking all the $x$ terms to the R.H.S. ;
$ \Rightarrow \sqrt y = \dfrac{{1 - x}}{{1 + x}}$
Taking square root on both the sides of the equation;
$\left( {\because \sqrt y {\text{ can also be written as }}{{\text{y}}^{\dfrac{1}{2}}}} \right)$
$ \Rightarrow {\left( {{y^{\dfrac{1}{2}}}} \right)^2} = {\left( {\dfrac{{1 - x}}{{1 + x}}} \right)^2}$
The above equation can also be written as;
$ \Rightarrow y = \dfrac{{{{\left( {1 - x} \right)}^2}}}{{{{\left( {1 + x} \right)}^2}}}{\text{ }}......\left( 1 \right)$
Now, differentiating equation $\left( 1 \right)$ with respect to $x$ , we get;
By the quotient or division rule of differentiation;
$ \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{\left( {\dfrac{{du}}{{dx}} \times v} \right) - \left( {u \times \dfrac{{dv}}{{dx}}} \right)}}{{{v^2}}}{\text{ }}......\left( 2 \right)$
Let $u = {\left( {1 - x} \right)^2}$ and $v = \left( {1 + {x^2}} \right)$
$ \Rightarrow \dfrac{{du}}{{dx}} = \dfrac{d}{{dx}}{\left( {1 - x} \right)^2}$
$ \Rightarrow \dfrac{{du}}{{dx}} = 2\left( {1 - x} \right) \times \left( { - 1} \right){\text{ }}......\left( 3 \right)$
Now, let us calculate $\dfrac{{dv}}{{dx}}$ ;
$ \Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{d}{{dx}}{\left( {1 + x} \right)^2}$
$ \Rightarrow \dfrac{{dv}}{{dx}} = 2\left( {1 + x} \right){\text{ }}......\left( 4 \right)$
Put the values of $\dfrac{{du}}{{dx}}{\text{ and }}\dfrac{{dv}}{{dx}}$ in equation $\left( 2 \right)$ , we get;
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{{{{\left( {1 - x} \right)}^2}}}{{{{\left( {1 + x} \right)}^2}}}} \right) = \dfrac{{\left( { - 2\left( {1 - x} \right) \times {{\left( {1 + x} \right)}^2}} \right) - \left( {{{\left( {1 - x} \right)}^2} \times 2\left( {1 + x} \right)} \right)}}{{{{\left( {{{\left( {1 + x} \right)}^2}} \right)}^2}}}$
Simplifying the above equation;
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2\left( {x - 1} \right) \times {{\left( {1 + x} \right)}^2} - {{\left( {x - 1} \right)}^2} \times 2\left( {1 + x} \right)}}{{{{\left( {1 + x} \right)}^4}}}$
Taking $2\left( {1 + x} \right)\left( {x - 1} \right)$ outside;
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2\left( {x - 1} \right)\left( {1 + x} \right)\left[ {\left( {1 + x} \right) - \left( {1 - x} \right)} \right]}}{{{{\left( {1 + x} \right)}^4}}}$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{2\left( {x - 1} \right)\left( {1 + x} \right)\left[ {\left( {1 + x - x + 1} \right)} \right]}}{{{{\left( {1 + x} \right)}^4}}}$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4\left( {x - 1} \right)\left( {1 + x} \right)}}{{{{\left( {1 + x} \right)}^4}}}$
$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{4\left( {x - 1} \right)}}{{{{\left( {1 + x} \right)}^3}}}$
Hence if $x = \left( {\dfrac{{1 - \sqrt y }}{{1 + \sqrt y }}} \right),$ then $\dfrac{{dy}}{{dx}} = \dfrac{{4\left( {x - 1} \right)}}{{{{\left( {1 + x} \right)}^3}}}$
Therefore, the correct answer for this question is option $\left( 2 \right)$.

Note: The knowledge of standard derivative formulae is very useful for solving such type of questions. Some of most important and frequently used derivative rules are given below:
$\left( 1 \right)$ The power rule : $\dfrac{d}{{dx}}{x^n} = n{x^{n - 1}}$ .
$\left( 2 \right)$ The addition rule: $\dfrac{d}{{dx}}\left( {u \pm v} \right) = \dfrac{{du}}{{dx}} \pm \dfrac{{dv}}{{dx}}$ .
$\left( 3 \right)$ Differentiation of a constant multiplied with a variable: $\dfrac{d}{{dx}}\left( {cx} \right) = c\dfrac{d}{{dx}}x = c$ .
$\left( 4 \right)$ The multiplication rule: $\dfrac{d}{{dx}}\left( {uv} \right) = v\dfrac{{du}}{{dx}} + u\dfrac{{dv}}{{dx}}$ .
$\left( 5 \right)$ The division or quotient rule $\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{\left( {\dfrac{{du}}{{dx}} \times v} \right) - \left( {u \times \dfrac{{dv}}{{dx}}} \right)}}{{{v^2}}}$ .