Question

Differentiate 2^2^2^x
Pls only students who passed 11th (PCM) answer it.

Answer 1

Answer:

2

1

+

2

x

ln

2

Step-by-step explanation:

Answer 2

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

$\rm :\longmapsto\:Let \: y = {2}^{ {2}^{ {2}^{x} } }$

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

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

$\rm :\longmapsto\:\dfrac{dy}{dx} = {2}^{ {2}^{ {2}^{x} } } log(2) \dfrac{d}{dx} {2}^{ \red{ {2}^{x}} }$

$\: \: \: \: \: \: \: \: \: \: \: \: \red{ \bigg \{ \because \: \dfrac{d}{dx} {a}^{x} = {a}^{x} log(a) \bigg \}}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {2}^{ {2}^{ {2}^{x} } } log(2) {2}^{{ {2}^{x}} } log(2)\dfrac{d}{dx} {2}^{x}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {2}^{ {2}^{ {2}^{x} } }{2}^{{ {2}^{x}} } {(log2)}^{2} \dfrac{d}{dx} {2}^{x}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {2}^{ {2}^{ {2}^{x} } } \: {2}^{{ {2}^{x}} } \: {(log2)}^{2} \: {2}^{x}log2$

$\bf :\longmapsto\:\dfrac{dy}{dx} = {2}^{ {2}^{ {2}^{x} } } \: {2}^{{ {2}^{x}} } \: {2}^{x} \: {(log2)}^{3}$

Alternative Method :-

$\rm :\longmapsto\:Let \: y = {2}^{ {2}^{ {2}^{x} } }$

On taking log both sides, we get

$\rm :\longmapsto\:log \: y = log \bigg({2}^{ {2}^{ {2}^{x} } } \bigg)$

$\bf :\longmapsto\:logy = {2}^{ {2}^{x} } log(2) - - - (1)$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \bigg \{ \because \: log( {x}^{y} ) = ylogx \bigg \}}$

Again on taking log both sides, we get

$\rm :\longmapsto\:log(logy) = log \bigg({2}^{ {2}^{x} } log(2) \bigg)$

$\rm :\longmapsto\:log(logy) = {2}^{x}log2 + log(log2)$

$\: \: \ \: \: \: \: \: \: \: \: \: \red{ \bigg \{ \because \: log( {x}^{y} ) = ylogx \bigg \}} \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \bigg \{ \because \: log(xy) = logx + logy\bigg \}}$

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

$\rm :\longmapsto\:\dfrac{d}{dx}log(logy) = \dfrac{d}{dx} \bigg({2}^{x}log2 + log(log2) \bigg)$

$\rm :\longmapsto\:\dfrac{1}{logy} \dfrac{d}{dx}logy = \dfrac{d}{dx} {2}^{x}log2 + \dfrac{d}{dx}log(log2)$

$\: \: \: \: \: \: \: \: \: \: \red{ \bigg \{ \because \: \dfrac{d}{dx}(u + v) = \dfrac{d}{dx}u + \dfrac{d}{dx}v \bigg \}}$

$\rm :\longmapsto\:\dfrac{1}{logy} \: \dfrac{1}{y} \: \dfrac{d}{dx}y = log2 \: \dfrac{d}{dx} {2}^{x} + 0$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \red{ \bigg \{ \because \:\dfrac{d}{dx}k = 0 \bigg \}}$

$\rm :\longmapsto\:\dfrac{1}{logy} \: \dfrac{1}{y} \: \dfrac{dy}{dx}= log2 \: ({2}^{x}log2)$

$\: \: \: \: \: \: \: \: \: \: \: \: \red{ \bigg \{ \because \: \dfrac{d}{dx} {a}^{x} = {a}^{x} log(a) \bigg \}}$

$\rm :\longmapsto\:\dfrac{1}{logy} \: \dfrac{1}{y} \: \dfrac{dy}{dx}= \: {2}^{x} {(log2)}^{2}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = y \: logy \: {2}^{x} \: {(log2)}^{2}$

$\rm :\longmapsto\:\dfrac{dy}{dx} = {2}^{ {2}^{ {2}^{x} } } \: {2}^{ {2}^{x} } \: log2 \: {2}^{x} \: {(log2)}^{2}$

$\bf :\longmapsto\:\dfrac{dy}{dx} = {2}^{ {2}^{ {2}^{x} } } \: {2}^{ {2}^{x} } \: {2}^{x} \: {(log2)}^{3}$

Additional Information :-

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

$\rm :\longmapsto\:\dfrac{d}{dx}k = 0$

$\rm :\longmapsto\:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}$

$\rm :\longmapsto\:\dfrac{d}{dx}k \: f(x) = k \: \dfrac{d}{dx}f(x)$

$\rm :\longmapsto\:\dfrac{d}{dx} {e}^{x} = {e}^{x}$

$\rm :\longmapsto\:\dfrac{d}{dx}logx = \dfrac{1}{x}$

$\rm :\longmapsto\:\dfrac{d}{dx} log_{a}(x) = \dfrac{1}{x \: loga} \: \: \: where \: a > 0$

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

$\rm :\longmapsto\:\dfrac{d}{dx}cosx = - \: sinx$

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

$\rm :\longmapsto\:\dfrac{d}{dx}cotx = - \: {cosec}^{2}x$