Question

The derivative of $ \cos e{c^{ - 1}}(\dfrac{1}{{2{x^2} - 1}}) $ with respect to $ \sqrt {1 - {x^2}} $ at $ x = \dfrac{1}{2} $ is:

Answer 1

Hint: Use the formulas of differentiation for trigonometric values and of their inverse, some of the used formulas in this Question is $ 2{\cos ^2}\theta - 1 = \cos 2\theta $ , $ 1 - {\cos ^2}\theta = {\sin ^2}\theta $ and many more, Put the appropriate values in the equations. And use of chain rule is important for these kinds of questions which ease the solving procedure.
Always prefer step by step solving rather than solving it an ounce.

Complete step-by-step answer:
Let $ u = $ $ \cos e{c^{ - 1}}(\dfrac{1}{{2{x^2} - 1}}) $ and $ v = $ $ \sqrt {1 - {x^2}} $
Let $ x = \cos \theta $ and put this in $ u $ and $ v $ : ………..(i)
In $ u = $ $ \cos e{c^{ - 1}}(\dfrac{1}{{2{x^2} - 1}}) $
Since, we know that $ 2{\cos ^2}\theta - 1 = \cos 2\theta $ :
So, $ u = $ $
  \cos e{c^{ - 1}}(\dfrac{1}{{\cos 2\theta }}) \\
  \dfrac{1}{{\cos 2\theta }} = \sec 2\theta \;
  $
Since, $ \dfrac{1}{{\cos 2\theta }} = \sec 2\theta $
So, $ u = $ $ \cos e{c^{ - 1}}(\sec 2\theta ) $
Differentiating $ u $ with respect to $ \theta $ using chain rule (In which $ \cos e{c^{ - 1}} $ is differentiated then $ \sec 2\theta $ and then $ 2\theta $ ), we get:
 $ \dfrac{{du}}{{d\theta }} = \dfrac{{ - 1}}{{{{\sec }^2}2\theta \sqrt {1 - \dfrac{1}{{{{\sec }^2}2\theta }}} }} \times \sec 2\theta \tan 2\theta \times 2
 $
 On further solving, with different formulas step by step we get:
So,
\[
  \dfrac{{du}}{{d\theta }} = \dfrac{{ - 1}}{{{{\sec }^2}2\theta \sqrt {\dfrac{{{{\sec }^2}2\theta - 1}}{{{{\sec }^2}2\theta }}} }} \times \sec 2\theta \tan 2\theta \times 2 \\
  \dfrac{{du}}{{d\theta }} = \dfrac{{ - 1}}{{{{\sec }^2}2\theta \sqrt {\dfrac{{{{\tan }^2}2\theta }}{{{{\sec }^2}2\theta }}} }} \times \sec 2\theta \tan 2\theta \times 2 \\
  \dfrac{{du}}{{d\theta }} = \dfrac{{ - 1}}{{{{\sec }^2}2\theta |\sqrt {\dfrac{{{{\tan }^2}2\theta }}{{{{\sec }^2}2\theta }}|} }} \times \sec 2\theta \tan 2\theta \times 2 \\
  \dfrac{{du}}{{d\theta }} = \dfrac{{ - 1}}{{{{\sec }^2}2\theta ( - \dfrac{{\tan 2\theta }}{{\sec 2\theta }})}} \times \sec 2\theta \tan 2\theta \times 2 \\
  \dfrac{{du}}{{d\theta }} = 2 \;
 \]
Put $ x = \cos \theta $ in $ v $ and solve it:
 $ v = $ $ \sqrt {1 - {x^2}} = \sqrt {1 - {{\cos }^2}\theta } $
Since, from the trigonometric identity we know that, $ 1 - {\cos ^2}\theta = {\sin ^2}\theta $
So, $ v = $ $ \sqrt {1 - {{\cos }^2}\theta } = \sqrt {{{\sin }^2}\theta } = \sin \theta $
Differentiating $ v $ with respect to $ \theta $ , we get:
 $ \dfrac{{dv}}{{d\theta }} = \cos \theta $ ………..(ii)
Put $ x = \dfrac{1}{2} $ in (i) and further solving using the trigonometric formulas, we get:
 $
  \dfrac{1}{2} = \cos \theta \\
  \cos \theta = \cos \dfrac{\pi }{3} \\
 \theta = \dfrac{\pi }{3} \\
  $
Put $ \theta = \dfrac{\pi }{3} $ in (ii), we get:
 $
  \dfrac{{dv}}{{d\theta }} = \cos \theta \\
  {(\dfrac{{dv}}{{d\theta }})_{\dfrac{\pi }{3}}} = \cos \dfrac{\pi }{3} = \dfrac{1}{2} \\
  $
Since, we had to find $ \dfrac{{du}}{{dv}} $ :
So, $ \dfrac{{du}}{{dv}} = \dfrac{{\dfrac{{du}}{{d\theta }}}}{{\dfrac{{dv}}{{d\theta }}}} $
Put the values obtained earlier in this equation and we get:
 $ \dfrac{{du}}{{dv}} = \dfrac{{\dfrac{{du}}{{d\theta }}}}{{\dfrac{{dv}}{{d\theta }}}} = \dfrac{2}{{\dfrac{1}{2}}} = \dfrac{{2 \times 2}}{1} = 4 $
So, $ {(\dfrac{{du}}{{dv}})_{\dfrac{1}{2}}} = 4 $
Therefore, the derivative of $ \cos e{c^{ - 1}}(\dfrac{1}{{2{x^2} - 1}}) $ with respect to $ \sqrt {1 - {x^2}} $ at $ x = \dfrac{1}{2} $ is: $ 4 $
So, the correct answer is “4”.

Note: Always remember the formula of differentiation of trigonometric values for easy solving.This contains various differentiation at various levels, so don’t get confused at different levels.At last always change or check the values at last before putting them into the last level.Remember the chain rule formula also. Any other formula would complicate the differentiation.