Question

$\underline{\sf{if \: y = { {x}^{x} }^{x} \: then \: find \: \frac{dy}{dx} }}$
Answer with proper explanation​

Answer 1

Answer:

We have,

y=(xx)x

y=xx2

 

On taking log both sides, we get

logy=x2logx                                               …… (1)

 

On differentiating w.r.t x, we get

y1dxdy=xx2+logx(2x)

y1dxdy=x+2xlogx

dxdy=y(x+2xlogx)

dxdy=(xx)x(x+2xlogx)

dxdy=x(xx)x(1+2logx)

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Answer 2

$\bf y = {x}^{{x}^{x} } \\ \\ \sf {\bf{ \underline{ \purple{Applying \: log \: in \: the \: both \: sides}}}} \\ \\ \sf log(y) = log({x}^{{x}^{x} }) \\ \\ \sf log(y) = {x}^{x} log(x) \\ \\{\bf{ \underline{ \purple{Differentiate \: w.r.t \: \: x}}}} \\ \\ \sf \dfrac{d log(y) }{dy}. \dfrac{dy}{dx} = {x}^{x}.\dfrac{d log(x) }{dx} + log(x) \dfrac{d ({x}^{x}) }{dx} \\ \\ \sf \dfrac{1}{y}.\dfrac{dy}{dx} = {x}^{x}. \dfrac{1}{x} + log(x) \dfrac{d ({x}^{x}) }{dx} \\ \\ { \bf{ \underline{ \purple{ Let \: {x}^{x} = v}}}} \\ \\ \sf {x}^{x} = v \\ \\ {\bf{ \underline{ \purple{Applying \: log \: in \: the \: both \: sides}}}} \\ \\ \sf log( {x}^{x}) = log(v) \\ \\ \sf log(v) = xlogx \\ \\ {\bf{ \underline{ \purple{Differentiate \: w.r.t \: \: x \: and \: applying \: product \: rule}}}} \\ \\ \sf \dfrac{dlog(v)}{dv} . \dfrac{dv}{dx} = x \dfrac{log(x)}{dx} + log(x). \dfrac{dx}{dx} \\ \\ \sf \frac{1}{v}. \dfrac{dv}{dx} = x. \frac{1}{x } + log(x) \\ \\ \sf \dfrac{dv}{dx} = {x}^{x}(1 + log(x)) \\ \\ {\bf{ \underline{ \purple{Now, \: substitute \: the \: value}}}} \\ \\ \sf \dfrac{1}{y}.\dfrac{dy}{dx} = {x}^{x}. \dfrac{1}{x} + log(x) \dfrac{d ({x}^{x}) }{dx} \\ \\ \sf \dfrac{1}{y}.\dfrac{dy}{dx} = {x}^{x}. \dfrac{1}{x} + log(x).{x}^{x} (1 + log(x)) \\ \\ \sf \dfrac{dy}{dx} = y{x}^{x}\[ \left [ \ \dfrac{1}{x} + log(x)(1 + log(x)) \right] \]$

• Quick review

Log pⁿ = n logp

Differentiation log p = 1/p

(uv)' = u'v + uv'