Question

if y= sec x + tan x show that the second derivative of y with respect to x is cos x/(1-sinx)²​

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\sf \:If \: y = secx + tanx, \: prove \: that \: y_2 = \dfrac{cosx}{ {(1 - sinx)}^{2} }$

Formula Used :-

$\boxed{\bf\:\dfrac{d}{dx}sinx = cosx}$

$\boxed{\bf\:\dfrac{d}{dx}cosx = - sinx}$

$\boxed{\bf\: \dfrac{d}{dx}k = 0}$

$\boxed{\bf\: \dfrac{d}{dx}y = y_1}$

$\boxed{\bf\: \dfrac{d}{dx}y_1 = y_2}$

$\boxed{\bf\: \dfrac{d}{dx}\dfrac{u}{v} = \dfrac{v\dfrac{d}{dx}u \: - \: u\dfrac{d}{dx}v}{ {v}^{2}}}$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:y = secx + tanx$

$\rm :\longmapsto\:y = \dfrac{1}{cosx} + \dfrac{sinx}{cosx}$

$\rm :\longmapsto\:y = \dfrac{1 + sinx}{cosx}$

On differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx} \dfrac{1 + sinx}{cosx}$

$\rm :\longmapsto\:y_1 = \dfrac{cosx\dfrac{d}{dx}(1 + sinx) - (1 + sinx)\dfrac{d}{dx}cosx}{ {cos}^{2}x }$

$\rm :\longmapsto\:y_1 = \dfrac{cosx(0 + cosx) - (1 + sinx)( - sinx)}{ {cos}^{2}x }$

$\rm :\longmapsto\:y_1 = \dfrac{ {cos}^{2}x + sinx + {sin}^{2}x}{ {cos}^{2}x}$

$\rm :\longmapsto\:y_1 = \dfrac{1 + sinx}{ {cos}^{2}x }$

$\rm :\longmapsto\:y_1 = \dfrac{1 + sinx}{1 - {sin}^{2}x }$

$\rm :\longmapsto\:y_1 = \dfrac{1 + sinx}{(1 - {sin}x)(1 + sinx) }$

$\bf\implies \: y_1= \dfrac{1}{1 - sinx}$

On differentiating both sides w. r. t. x, we get

$\bf\implies \: \dfrac{d}{dx}y_1=\dfrac{d}{dx} \dfrac{1}{1 - sinx}$

$\rm :\longmapsto\:y_2 = \dfrac{d}{dx} {(1 - sinx)}^{ - 1}$

$\rm :\longmapsto\:y_2 = - 1 {(1 - sinx)}^{ - 1 - 1}\dfrac{d}{dx}(1 - sinx)$

$\rm :\longmapsto\:y_2 = - 1 {(1 - sinx)}^{ - 2}(0 - cosx)$

$\bf\implies \:y_2 = \dfrac{cosx}{ {(1 - sinx)}^{2} }$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

Additional Information :-

$\boxed{\bf\: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}$

$\boxed{\bf\: \dfrac{d}{dx}x = 1}$

$\boxed{\bf\: \dfrac{d}{dx}tanx = {sec}^{2}x}$

$\boxed{\bf\: \dfrac{d}{dx}cotx = { - \: cosec}^{2}x}$

$\boxed{\bf\: \dfrac{d}{dx}secx = secx \: tanx}$

$\boxed{\bf\: \dfrac{d}{dx}cosecx = - \: cosecx \: cotx}$