Question

If y = (loge x)^x + x^logex. find
dy/dx​

Answer 1

Solution:-

$\implies \rm \: y = ( log_{e}x) {}^{x} + {x}^{ log_{e}x}$

To find

$\rm \implies \dfrac{dy}{dx}$

Now

$\implies \rm \: y = ( log_{e}x) {}^{x} + {x}^{ log_{e}x}$

Let

$\implies \rm \: a= ( log_{e}x) {}^{x} \: \: and \: \: \: b = {x}^{ log_{e}x}$

$\rm \implies \: y = a + b$

Take

$\rm \implies \: a = ( log_{e}x) {}^{x}$

Taking both side log

$\rm \implies log_{e}a = log( log_{e}x) {}^{x}$

Using logarithmic properties

$\rm \implies log_{e}a=x log( log_{e}x)$

Now differentiate w.r.t x

$\rm \implies \dfrac{1}{a} \times \dfrac{da}{dx} = x \times \dfrac{ 1 }{ log_{e}x} \times \dfrac{1}{x} + log( log_{e}x) \times 1$

In above step we use product rule of differentiation

$\rm \implies \dfrac{1}{a} \times \dfrac{da}{dx} = \cancel{ x }\times \dfrac{ 1 }{ log_{e}x} \times \dfrac{1}{ \cancel{x}} + log( log_{e}x) \times 1$

$\rm \implies \dfrac{da}{dx} = \bigg \{ \dfrac{ 1 }{ log_{e}x} + log( log_{e}x) \bigg \} \times a$

So value of a is

$\rm \implies \: a = ( log_{e}x) {}^{x}$

By putting the value

$\rm \implies \dfrac{da}{dx} = \bigg \{ \dfrac{ 1 }{ log_{e}x} + log( log_{e}x) \bigg \} ( log_{e}x) {}^{x}$

Now take

$\rm \implies \: {x}^{ log_{e}x } = b$

Taking both side log

$\implies \rm log(x {}^{ log_{e}x} ) = log b$

Using logarithmic properties

$\rm \implies logx \times logx = logb$

$\implies \rm logb = (logx) ^{2}$

Differentiate wrt to x

$\rm \implies \: \dfrac{1}{b} \times \dfrac{db}{dx} = 2 logx \times \dfrac{1}{x}$

In above step using chain rule

$\rm \implies \: \dfrac{db}{dx} = \bigg \{2 logx \times \dfrac{1}{x} \bigg \} \times b$

Put the value of b

$\rm \implies \: {x}^{ log_{e}x } = b$

we get

$\rm \implies \: \dfrac{db}{dx} = \bigg \{2 logx \times \dfrac{1}{x} \bigg \} \times {x}^{ log_{e}x }$

Now take

$\rm \implies \: y = a + b$

Differentiate with respect to x

$\rm \implies \: \dfrac{dy}{dx} = \dfrac{da}{dx} + \dfrac{db}{dx}$

So put the value of da/dx and db/dx

$\rm \implies \: \dfrac{dy}{dx} = \bigg \{ \dfrac{ 1 }{ log_{e}x} + log( log_{e}x) \bigg \} ( log_{e}x) {}^{x} + \bigg \{2 logx \times \dfrac{1}{x} \bigg \} \times {x}^{ log_{e}x }$

Answer

$\rm \implies \: \dfrac{dy}{dx} = \bigg \{ \dfrac{ 1 }{ log_{e}x} + log( log_{e}x) \bigg \} ( log_{e}x) {}^{x} + \bigg \{2 logx \times \dfrac{1}{x} \bigg \} \times {x}^{ log_{e}x }$