Question

Find the value of \[\dfrac{{dy}}{{dx}}\] ,If \[x = y\sqrt {\left( {1 - {y^2}} \right)} \]
A. \[x\]
B.\[\sqrt {\dfrac{{\left( {1 - {y^2}} \right)}}{{\left( {1 + 2{y^2}} \right)}}} \]
C. \[0\]
D.\[\dfrac{{\sqrt {1 - {y^2}} }}{{1 - 2{y^2}}}\]

Answer 1

Hint:
We will be directly Differentiating the equation given as \[x = y\sqrt {\left( {1 - {y^2}} \right)} \]with respect to \[dx\] .

Formula Used:
We will be using product rule \[\dfrac{d}{{dx}}[f(x) \cdot g(x)] = \dfrac{d}{{dx}}[f(x)] \cdot g(x) + f(x) \cdot \dfrac{d}{{dx}}[g(x)]\] to differentiate the equation .


Complete step-by-step answer:
Integrate the term \[x = y\sqrt {\left( {1 - {y^2}} \right)} \] with respect to \[dx\].
\[x = y\sqrt {\left( {1 - {y^2}} \right)} \]
\[\dfrac{{dx}}{{dx}} = \dfrac{d}{{dx}}\left( {y\sqrt {1 - {y^2}} } \right) \times \dfrac{{dy}}{{dy}}\]
\[1 = \dfrac{d}{{dy}}\left( {y\sqrt {1 - {y^2}} } \right).\dfrac{{dy}}{{dx}}\]
Now by applying Product Rule \[\dfrac{d}{{dx}}[f(x) \cdot g(x)] = \dfrac{d}{{dx}}[f(x)] \cdot g(x) + f(x) \cdot \dfrac{d}{{dx}}[g(x)]\]
\[1 = \dfrac{d}{{dy}}\left( {y\sqrt {1 - {y^2}} } \right).\dfrac{{dy}}{{dx}}\]
\[\left( {\sqrt {1 - {y^2}} + y\dfrac{1}{{2\sqrt {1 - {y^2}} }} \times \left( { - 2y} \right)} \right)\dfrac{{dy}}{{dx}} = 1\]
\[\left( {\sqrt {1 - {y^2}} - \dfrac{{{y^2}}}{{\sqrt {1 - {y^2}} }}} \right)\dfrac{{dy}}{{dx}} = 1\]
\[\left( {\dfrac{{1 - {y^2} - {y^2}}}{{\sqrt {1 - {y^2}} }}} \right)\dfrac{{dy}}{{dx}} = 1\]
\[\left( {\dfrac{{1 - 2{y^2}}}{{\sqrt {1 - {y^2}} }}} \right)\dfrac{{dy}}{{dx}} = 1\]
\[\dfrac{{dy}}{{dx}} = \dfrac{{\sqrt {1 - {y^2}} }}{{1 - 2{y^2}}}\]
Hence the answer is (D) which is\[\dfrac{{dy}}{{dx}} = \dfrac{{\sqrt {1 - {y^2}} }}{{1 - 2{y^2}}}\].



Note:
Students are often confused with chain rule and product rule. The chain rule is applicable when we have to find the derivative of a composite function. The product rule is applicable when two functions multiply with each other. In the given question two functions are multiplied by each other. Thus we apply the product rule.