Question

y=log(secx) find d⁴y/dx⁴​

Answer 1

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

$\rm :\longmapsto\:y = log(secx)$

$\large\underline{\sf{To\:Find - }}$

$\boxed{ \sf{\rm :\longmapsto\: \:\dfrac{ {d}^{4} y}{ {dx}^{4} }}}$

$\begin{gathered}\Large{ \sf{{\underline{Formula \: Used - }}}}  \end{gathered}$

$\underbrace{\boxed{ \tt{ \dfrac{d}{dx}logx = \frac{1}{x} \: }}}$

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

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

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

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

Given that

$\rm :\longmapsto\:y = log(secx)$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y =\dfrac{d}{dx} log(secx)$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{secx}\dfrac{d}{dx}secx$

$\rm :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{secx}(secx \: tanx)$

$\bf\implies \:\dfrac{dy}{dx} = tanx$

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

$\rm :\longmapsto\: \:\dfrac{d}{dx}\dfrac{dy}{dx} = \dfrac{d}{dx}tanx$

$\rm :\longmapsto\:\dfrac{ {d}^{2}y }{d {x}^{2} } = {sec}^{2}x$

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

$\rm :\longmapsto\:\dfrac{d}{dx}\dfrac{ {d}^{2}y }{d {x}^{2} } = \dfrac{d}{dx}{sec}^{2}x$

$\rm :\longmapsto\:\dfrac{ {d}^{3}y }{d {x}^{3} } = 2{sec}x \: \dfrac{d}{dx}secx$

$\rm :\longmapsto\:\dfrac{ {d}^{3}y }{d {x}^{3} } = 2{sec}x \: (secx \: tanx)$

$\rm :\longmapsto\:\dfrac{ {d}^{3}y }{d {x}^{3} } = 2 \: {sec}^{2}x \: tanx$

We know,

$\underbrace{\boxed{ \tt{ {sec}^{2}x \: = \: 1 + {tan}^{2}x }}}$

So, using this

$\rm :\longmapsto\:\dfrac{ {d}^{3}y }{d {x}^{3} } = 2 \: (1 + {tan}^{2}x) \: tanx$

$\rm :\longmapsto\:\dfrac{ {d}^{3}y }{d {x}^{3} } = 2 \: (tanx + {tan}^{3}x)$

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

$\rm :\longmapsto\:\dfrac{d}{dx}\dfrac{ {d}^{3}y }{d {x}^{3} } = \dfrac{d}{dx} 2 \: (tanx + {tan}^{3}x)$

$\rm :\longmapsto\:\dfrac{ {d}^{4}y }{d {x}^{4} } = 2 \: (\dfrac{d}{dx}tanx + \dfrac{d}{dx}{tan}^{3}x)$

$\rm :\longmapsto\:\dfrac{ {d}^{4}y }{d {x}^{4} } = 2 \: \bigg( {sec}^{2}x + 3 {tan}^{2}x\dfrac{d}{dx}tanx \bigg)$

$\rm :\longmapsto\:\dfrac{ {d}^{4}y }{d {x}^{4} } = 2 \: \bigg( {sec}^{2}x + 3 {tan}^{2}x {sec}^{2}x \bigg)$

$\red{ \boxed{\rm :\longmapsto\:\dfrac{ {d}^{4}y }{d {x}^{4} } = 2 \: {sec}^{2}x \: \bigg( 1 + 3 {tan}^{2}x\bigg) \quad}}$

or

in further evaluation we get

We know,

$\underbrace{\boxed{ \tt{ {sec}^{2}x \: - \: {tan}^{2}x \: = \: 1}}}$

So, using this

$\rm :\longmapsto\:\dfrac{ {d}^{4}y }{d {x}^{4} } = 2 \: {sec}^{2}x \: \bigg( {sec}^{2}x - {tan}^{2}x + 3 {tan}^{2}x\bigg)$

$\rm :\longmapsto\:\dfrac{ {d}^{4}y }{d {x}^{4} } = 2 \: {sec}^{2}x \: \bigg( {sec}^{2}x + 2 {tan}^{2}x\bigg)$

$\red{ \boxed{\rm :\longmapsto\:\dfrac{ {d}^{4}y }{d {x}^{4} } = 2 \: {sec}^{4}x \: +4\: {sec}^{2}x \: {tan}^{2}x \quad}}$

Additional Information :-

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

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

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

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

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

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