Question

Find $\dfrac{{dy}}{{dx}}$ , if $y = {\sin ^{ - 1}}\left( {\dfrac{{6x - 4\sqrt {1 - 4{x^2}} }}{5}} \right)$ .

Answer 1

Hint : Use the inverse trigonometric differentiation formulas to solve the given question. Also differentiate the angle of the trigonometric property given.

Complete step-by-step answer :
As we know that is ${\sin ^{ - 1}}x = y$ , then it is also true that $\sin y = x$ . So, simplify the given expression with this rule.
Now simplify the given trigonometric identity as given below:
$y = {\sin ^{ - 1}}\left( {\dfrac{{6x - 4\sqrt {1 - 4{x^2}} }}{5}} \right) \\\ \sin y = \dfrac{{6x - 4\sqrt {1 - 4{x^2}} }}{5} \\\$
The chain rule of differentiation says that is a function $h\left( x \right)$ can be written as $h\left( x \right) = f\left( {g\left( x \right)} \right)$ , then $\dfrac{d}{{dx}}\left( {h\left( x \right)} \right)$ will be equal to $f'\left( {g\left( x \right)} \right) \times \dfrac{d}{{dx}}\left( {g\left( x \right)} \right)$ .
Now differentiate both the sides with respect to $x$ using the chain rule:
$\dfrac{d}{{dx}}\left( {\sin y} \right) = \dfrac{d}{{dx}}\left( {\dfrac{{6x - 4\sqrt {1 - 4{x^2}} }}{5}} \right) \\\ \cos y\dfrac{{dy}}{{dx}} = \dfrac{6}{5}\dfrac{d}{{dx}}\left( x \right) - \dfrac{4}{5}\dfrac{d}{{dx}}\left( {\sqrt {1 - 4{x^2}} } \right) \\\ \sqrt {\left( {1 - {{\sin }^2}y} \right)} \dfrac{{dy}}{{dx}} = \dfrac{6}{5} - \dfrac{4}{5} \times \dfrac{1}{{2\sqrt {1 - 4{x^2}} }} \times \dfrac{d}{{dx}}\left( {1 - 4{x^2}} \right) \\\ \sqrt {\left( {1 - {{\left( {\dfrac{{6x - 4\sqrt {1 - 4{x^2}} }}{5}} \right)}^2}} \right)} \dfrac{{dy}}{{dx}} = \dfrac{6}{5} - \dfrac{4}{5} \times \dfrac{{ - 8x}}{{2\sqrt {1 - 4{x^2}} }} \\\$
Further simplify,
$\sqrt {\left( {1 - {{\left( {\dfrac{{6x - 4\sqrt {1 - 4{x^2}} }}{5}} \right)}^2}} \right)} \dfrac{{dy}}{{dx}} = \dfrac{6}{5} - \dfrac{4}{5} \times \dfrac{{ - 8x}}{{2\sqrt {1 - 4{x^2}} }} \\\ \sqrt {\left( {1 - \dfrac{{16 - 28{x^2} - 48x\sqrt {1 - 4{x^2}} }}{{25}}} \right)} \dfrac{{dy}}{{dx}} = \dfrac{1}{5}\left( {6 + \dfrac{{16x}}{{\sqrt {1 - 4{x^2}} }}} \right) \\\ \sqrt {\left( {\dfrac{{28{x^2} + 48x\sqrt {1 - 4{x^2}} + 9}}{{25}}} \right)} \dfrac{{dy}}{{dx}} = \dfrac{1}{5}\left( {\dfrac{{6\sqrt {1 - 4{x^2}} + 16x}}{{\sqrt {1 - 4{x^2}} }}} \right) \\\ \dfrac{{dy}}{{dx}} = \dfrac{{\left( {6\sqrt {1 - 4{x^2}} + 16x} \right)}}{{\sqrt {1 - 4{x^2}} \left( {\sqrt {28{x^2} + 48x\sqrt {1 - 4{x^2}} + 9} } \right)}} \\\$
Hence the value of $\dfrac{{dy}}{{dx}}$ is equal to $\dfrac{{\left( {6\sqrt {1 - 4{x^2}} + 16x} \right)}}{{\sqrt {1 - 4{x^2}} \left( {\sqrt {28{x^2} + 48x\sqrt {1 - 4{x^2}} + 9} } \right)}}$ for the given question.
So, the correct answer is “$\dfrac{{\left( {6\sqrt {1 - 4{x^2}} + 16x} \right)}}{{\sqrt {1 - 4{x^2}} \left( {\sqrt {28{x^2} + 48x\sqrt {1 - 4{x^2}} + 9} } \right)}}$ ”.

Note : Use the chain rule to differentiate the term $\sin y$ with respect to the variable $x$ as the variable $y$ is also a function of $x$ itself. The trigonometric identity ${\sin ^2}x + {\cos ^2}x = 1$ holds true. The chain rule of differentiation says that is a function $h\left( x \right)$ can be written as $h\left( x \right) = f\left( {g\left( x \right)} \right)$ , then $\dfrac{d}{{dx}}\left( {h\left( x \right)} \right)$ will be equal to $f'\left( {g\left( x \right)} \right) \times \dfrac{d}{{dx}}\left( {g\left( x \right)} \right)$ .