Question

if y=1/5 log (5+tan x/2/5-tan x/2),prove that dy/dx= 1/12+13+cosx​

Answer 1

Identities Used :-

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

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

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

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

$\boxed{ \sf{ \:(x + y)(x - y) = {x}^{2} - {y}^{2} }}$

$\boxed{ \sf{ \: tanx = \dfrac{sinx}{cosx} }}$

$\boxed{ \sf{ \:1 + cos2x = {2cos}^{2}x}}$

$\boxed{ \sf{ \:1 - cos2x = {2sin}^{2}x}}$

Lets Solve the problem now!!!

★ Given

$\rm :\longmapsto\:y = \dfrac{1}{5} log\bigg(\dfrac{5 + tan\dfrac{x}{2} }{5 - tan\dfrac{x}{2}} \bigg)$

★ We know,

$\boxed{ \sf \: log \frac{x}{y} = logx \: - \: logy}$

So,

★ Given expression can be rewritten as

$\rm :\longmapsto\:y = \dfrac{1}{5} log\bigg(5 + tan\dfrac{x}{2}\bigg) - \dfrac{1}{5}log\bigg(5 - tan\dfrac{x}{2} \bigg)$

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

$\rm :\longmapsto\:\dfrac{d}{dx}y = \dfrac{1}{5} \dfrac{d}{dx}log\bigg(5 + tan\dfrac{x}{2}\bigg)-\dfrac{1}{5} \dfrac{d}{dx}log\bigg(5 - tan\dfrac{x}{2} \bigg)$

$\rm \:=\dfrac{1}{5} \bigg(\dfrac{1}{5 + tan\dfrac{x}{2}}\dfrac{d}{dx}(5 + tan\dfrac{x}{2}) - \dfrac{1}{5-tan\dfrac{x}{2}}\dfrac{d}{dx}(5 - tan\dfrac{x}{2}) \bigg)$

$\rm \:=\dfrac{1}{5} \bigg(\dfrac{1}{5 + tan\dfrac{x}{2}} \: {sec}^{2} \dfrac{x}{2} \times \dfrac{1}{2} + \dfrac{1}{5-tan\dfrac{x}{2}} {sec}^{2} \dfrac{x}{2} \times\dfrac{1}{2} \bigg)$

$\rm \:= \dfrac{1}{10} {sec}^{2}\dfrac{x}{2} \bigg(\dfrac{1}{5 + tan\dfrac{x}{2}} \: + \dfrac{1}{5-tan\dfrac{x}{2}} \bigg)$

$\rm \:= \dfrac{1}{10} {sec}^{2}\dfrac{x}{2} \bigg(\dfrac{5 - tan\dfrac{x}{2} + 5 + tan\dfrac{x}{2}}{(5 + tan\dfrac{x}{2})(5 - tan\dfrac{x}{2})} \: \bigg)$

$\rm \:= \dfrac{1}{10} {sec}^{2}\dfrac{x}{2} \bigg(\dfrac{10}{25 - {tan}^{2} \dfrac{x}{2}} \: \bigg)$

$\rm \:= {sec}^{2}\dfrac{x}{2} \bigg(\dfrac{1}{25 - {tan}^{2} \dfrac{x}{2}} \: \bigg)$

$\rm \:= {sec}^{2}\dfrac{x}{2} \bigg(\dfrac{ {cos}^{2}\dfrac{x}{2} }{25 {cos}^{2}\dfrac{x}{2} - {sin}^{2} \dfrac{x}{2}} \: \bigg)$

$\rm \:= \bigg(\dfrac{ 1 }{25 {cos}^{2}\dfrac{x}{2} - {sin}^{2} \dfrac{x}{2}} \: \bigg)$

$\rm \:= \bigg(\dfrac{ 1 }{25 \bigg(\dfrac{1 + cosx}{2} \bigg) - \dfrac{1 - cosx}{2}} \: \bigg)$

$\rm \:= \bigg(\dfrac{2 }{25 + 25cosx - 1 + cosx} \: \bigg)$

$\rm \:= \bigg(\dfrac{2 }{24 + 26cosx} \: \bigg)$

$\rm \:= \bigg(\dfrac{2 }{2(12 + 13cosx)} \: \bigg)$

$\rm \:= \bigg(\dfrac{1 }{12 + 13cosx} \:\bigg)$

Hence,

$\bf :\longmapsto\:\dfrac{dy}{dx} = \dfrac{1}{12 + 13 \: cosx}$