"Let $$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$, 0

Question

"Let $$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$, 0 < x < 1. Then :"

Accepted Answer

"$$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$

${\cot ^{ - 1}}\sqrt {{{1 - x} \over x}} = {\sin ^{ - 1}}\sqrt x$

or $f(x) = \cos (2{\tan ^{ - 1}}\sqrt x )$

$= \cos {\tan ^{ - 1}}\left( {{{2\sqrt x } \over {1 - x}}} \right)$

$f(x) = {{1 - x} \over {1 + x}}$

Now, $f'(x) = {{ - 2} \over {{{(1 + x)}^2}}}$

or $f'(x){(1 - x)^2} = - 2{\left( {{{1 - x} \over {1 + x}}} \right)^2}$

or ${(1 - x)^2}f'(x) + 2{(f(x))^2} = 0$."

Let $$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$, 0 < x < 1. Then :

Solution

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Let $$f(x) = \cos \left( {2{{\tan }^{ - 1}}\sin \left( {{{\cot }^{ - 1}}\sqrt {{{1 - x} \over x}} } \right)} \right)$$, 0 < x < 1. Then :

Solution

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