Question

pls answer this question ASAP

Answer 1

Answer

• f'(x) = ¹/₁ ₊ cos x
• f'(π/₃) = ²/₃

Given

$\bullet \;\;\; \rm f(x)=\sqrt{\dfrac{\sec x-1}{\sec x+1}}$

To Find

$\bullet \;\;\; \rm f'(x)$

Solution

$\rm \sqrt{\dfrac{sec\ x-1}{sec\ x+1}}\\\\\rm On\ Rationalizing\ the\ numerator\ ,we\ get ,\\\\\implies \rm \sqrt{\dfrac{sec\ x-1}{sec\ x+1}\times \dfrac{sec\ x+1}{sec\ x+1}}\\\\\implies \rm \sqrt{\dfrac{sec^2x-1^2}{(sec\ x+1)^2}}\\\\\implies \rm \sqrt{\dfrac{sec^2x-1}{(sec\ x+1)^2}}\\\\\implies \rm \sqrt{\dfrac{tan^2x}{(sec\ x+1)^2}}\\\\\implies \rm \dfrac{\sqrt{tan^2x}}{\sqrt{(sec\ x+1)^2}}\\\\\implies \rm \dfrac{tan\ x}{sec\ x+1}$

So ,

$\rm f(x)=\dfrac{tan\ x}{sec\ x+1}$

Now , we need to find f'(x) which is nothing but derivative .

$\rm Let,y=f(x)\\\\\implies \rm y=\dfrac{tan\ x}{sec\ x+1}$

Differentiating with respect to "x" on both sides , we get ,

$\implies \rm \dfrac{dy}{dx}=\dfrac{d}{dx}\bigg( \dfrac{tan\ x}{sec\ x+1}\bigg)\\\\\implies \rm \dfrac{dy}{dx}=\dfrac{(sec\ x+1)\frac{d}{dx}(tan\ x)-tan\ x\frac{d}{dx}(sec\ x+1)}{(sec\ x+1)^2}\\\\\implies \rm \dfrac{dy}{dx}=\dfrac{(sec\ x+1)sec^2x-tan\ x(sec\ x.tan\ x)}{(sec\ x+1)^2}\\\\\implies \rm \dfrac{dy}{dx}=\dfrac{sec^3x+sec^2x-sec\ x.tan^2x}{(sec\ x+1)^2}\\\\\implies \rm \dfrac{dy}{dx}=\dfrac{sec^3x+sec^2x-sec\ x(sec^2x-1)}{(sec\ x+1)^2}$

$\implies \rm \dfrac{dy}{dx}=\dfrac{sec^3x+sec^2x-sec^3x+secx}{(sec\ x+1)^2}\\\\\implies \rm \dfrac{dy}{dx}=\dfrac{sec^2x+sec\ x}{(sec\ x+1)^2}\\\\\implies \rm \dfrac{dy}{dx}=\dfrac{sec\ x(sec\ x+1)}{(sec\ x+1)^2}\\\\\implies \rm \dfrac{dy}{dx}=\dfrac{sec\ x}{sec\ x+1}\\\\\implies \rm \dfrac{dy}{dx}=\dfrac{\frac{1}{cos\ x}}{\frac{1}{cos\ x}+1}\\\\\implies \rm \dfrac{dy}{dx}=\dfrac{\frac{1}{cos\ x}}{\frac{1+cos\ x}{cos\ x} }\\\\\implies\rm \dfrac{dy}{dx}=\dfrac{1}{cos\ x}\times\dfrac{cos\ x}{1+cos\ x}$

$\implies \rm \dfrac{dy}{dx}=\dfrac{1}{1+cos\ x}$

$\rm So,f'(x)=\dfrac{dy}{dx}=\dfrac{1}{1+cos\ x}\\\\\implies \bf f'(x)=\dfrac{1}{1+cos\ x}\\\\\implies \rm f'\bigg(\dfrac{\pi}{3}\bigg)=\dfrac{1}{1+cos\ \bigg( \dfrac{\pi}{3}\bigg)}\\\\\implies \rm f'\bigg( \dfrac{\pi}{3}\bigg)=\dfrac{1}{1+cos\ 60^0}\\\\\implies \rm f'\bigg(\dfrac{\pi}{3}\bigg)=\dfrac{1}{1+\frac{1}{2}}\\\\\implies \rm f'\bigg(\dfrac{\pi}{3}\bigg)=\dfrac{1}{\frac{2+1}{2}}\\\\\implies \rm f'\bigg(\dfrac{\pi}{3}\bigg)=\dfrac{1}{\frac{3}{2}}$

$\implies \bf f'\bigg( \dfrac{\pi}{3}\bigg)=\dfrac{2}{3}$

Formula Used

$\bullet \;\; \bf sec^2x-tan^2x=1\\\\\bullet \;\; \dfrac{d}{dx}(tan\ x)=sec^2x\\\\\bullet \;\; \dfrac{d}{dx}(sec\ x)=sec\ x.tan\ x$