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) $ .