Question

If y=log root over 1+tanx/1-tanx, prove that dy/dx =sec2x

Answer 1

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Answer 2

∴ proved that  $\frac{dy}{dx} =sec2x$

Step-by-step explanation:

Given;

$y=log\sqrt{\frac{1+tanx}{1-tanx} }$

Differentiate above equation with respect to $x$,

  $\frac{dy}{dx} =\frac{1}{dx} log\sqrt{\frac{1+tanx}{1-tanx} }$

$\frac{dy}{dx} =\frac{1}{\sqrt{\frac{1+tanx}{1-tanx} } } \frac{1}{dx}\sqrt{\frac{1+tanx}{1-tanx} }$

$\frac{dy}{dx} =\frac{1}{\sqrt{\frac{1+tanx}{1-tanx} } }\times\frac{1}{2\sqrt{\frac{1+tanx}{1-tanx} } }\times \frac{2sec^{2}x }{(1-tanx)^{2} }$

$\frac{dy}{dx} =\frac{1}{2} \sqrt{\frac{1-tanx}{1+tanx} } \times \sqrt{\frac{1-tanx}{1+tanx} } \times \frac{2sec^{2}x }{(1-tanx)^{2} }$

$\frac{dy}{dx} =\frac{1}{2} \times\frac{1-tanx}{1+tanx} } \times \frac{2sec^{2}x }{(1-tanx)^{2} }$

$\frac{dy}{dx} =\frac{1}{1+tanx} } \times \frac{sec^{2}x }{(1-tanx) }$

$\frac{dy}{dx} =\frac{sec^{2}x }{(1-tan^{2} x) }$

$\frac{dy}{dx} =\frac{\frac{1}{cos^{2}x}} {(1-\frac{sin^{2}x }{cos^{2}x } ) }$

$\frac{dy}{dx} =\frac{1}{cos^{2}x-sin^{2} x }$

$\frac{dy}{dx} =\frac{1}{cos2x}$

$\frac{dy}{dx} =sec2x$

So proved that  $\frac{dy}{dx} =sec2x$