Question

If $y = \dfrac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}$, prove that $(1 - {x^2})\dfrac{{dy}}{{dx}} = (xy + 1)$

Answer 1

In the question they have given the function $y$ , and we know that if $y = \dfrac{u}{v}$ where $u,v$ are the functions dependent on $x$. If we differentiate the function $y$ with respect to $x$ that is $\dfrac{{dy}}{{dx}}$ we get $\dfrac{{dy}}{{dx}} = \dfrac{{vdu - udv}}{{{v^2}}}$ .By using this formula if we differentiate and simplify the given function we arrive at the correct solution.

Complete step-by-step solution:
In the question, we have given that
$y = \dfrac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}$
And we have to prove that $(1 - {x^2})\dfrac{{dy}}{{dx}} = (xy + 1)$
So first of all we need to find the value of $\dfrac{{dy}}{{dx}}$
And also we know for any function in the form of $y = \dfrac{u}{v}$ where $u,v$ are the functions dependent on $x$ and then $\dfrac{{dy}}{{dx}}$ is given by the formula
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{vdu - udv}}{{{v^2}}}$
So here we have given that $u = {\sin ^{ - 1}}x$ and $v = \sqrt {1 - {x^2}}$ as in the question.
So we need to find the value of $du,dv$
So we have $u = {\sin ^{ - 1}}x$
$\sin u = x$
Now upon differentiating the above functions with respect of $x$, we get
$\cos u\left( {\dfrac{{du}}{{dx}}} \right) = 1$
$\left( {\dfrac{{du}}{{dx}}} \right) = \dfrac{1}{{\cos u}}$
And we also know that
${\sin ^2}u + {\cos ^2}u = 1$
$\cos u = \sqrt {1 - {{\sin }^2}u}$
$\Rightarrow$ $\dfrac{{du}}{{dx}} = \dfrac{1}{{\sqrt {1 - {{\sin }^2}u} }}$
And we have written above, that is from the question
$\sin u = x$
So $\dfrac{{du}}{{dx}} = \dfrac{1}{{\sqrt {1 - {x^2}} }}$
$\Rightarrow$ $du = \dfrac{1}{{\sqrt {1 - {x^2}} }}dx$
Now let we assume that
$v = \sqrt {1 - {x^2}}$
$\Rightarrow$ $\dfrac{{dv}}{{dx}} = \dfrac{1}{{2\sqrt {1 - {x^2}} }}. - 2x$ $\Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{{ - xdx}}{{\sqrt {1 - {x^2}} }}$
Now we know by previous discussion that $y = \dfrac{u}{v}$ where $u,v$ are the function dependent on $x$ and then $\dfrac{{dy}}{{dx}}$ is given by $\dfrac{{dy}}{{dx}} = \dfrac{{vdu - udv}}{{{v^2}}}$ $- - - (1)$
And as we know that $v = \sqrt {1 - {x^2}}$ $\Rightarrow \dfrac{{dv}}{{dx}} = \dfrac{{ - xdx}}{{\sqrt {1 - {x^2}} }}$
$\Rightarrow$ $u = {\sin ^{ - 1}}x$,$\dfrac{{du}}{{dx}} = \dfrac{1}{{\sqrt {1 - {x^2}} }}$
Putting it in equation (1)
$\Rightarrow$ $\dfrac{{dy}}{{dx}} = \dfrac{{\sqrt {1 - {x^2}} \times \dfrac{1}{{\sqrt {1 - {x^2}} }} - {{\sin }^{ - 1}}x \times \dfrac{{ - 2x}}{{2\sqrt {1 - {x^2}} }}}}{{{{(\sqrt {1 - {x^2}} )}^2}}}$
$\Rightarrow$ $\dfrac{{dy}}{{dx}} = \dfrac{{1 + \dfrac{{x{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}}}{{1 - {x^2}}}$
By cross multiplication, we will get
$\Rightarrow$ $\dfrac{{dy}}{{dx}}(1 - {x^2}) = 1 + \dfrac{{x{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}$
And we know that $y = \dfrac{{{{\sin }^{ - 1}}x}}{{\sqrt {1 - {x^2}} }}$
So we get $\dfrac{{dy}}{{dx}}(1 - {x^2}) = 1 + xy$
Hence proved

Note: Whenever we have a function given that is $y$, first we need to differentiate that function with respect to the corresponding function that is $x$, if the answer you get is not what they have asked for then try to differentiate it again to arrive at the solution. For any given function $y = g(f(x))$ its derivative will be given by $\dfrac{{dy}}{{dx}} = g'(f(x))f'(x)$.