Question

Find dy/dx , y = (x^x)^x​

Answer 1

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Actually Welcome to the Concept of the Derivatives by use of the Logarithms

Now basically here we are going to use the Logarithms to take derivatives that are in the index form

so here we are given with,

let's get here first, derivative of y= x^x

so we get as, log y = xlogx

=> 1/y dy/dx = x. (1/x) + logx

=> dy/dx = y. (1+logx)

that is, dy/dx = x^x(1+logx)......... (1)

so now we move forward

=> y = (x^x)^x

so taking log on both sides,

=> log y = xlog(x^x)

now, differentiating by using the product rule and the chain rule we get as,

from... (1) we get as,

=> 1/y dy/dx = x. 1/(x^x).(x^x) (1+logx) + log(x^x). (1)

now again solving we get as,

=> 1/y dy/dx = x. (1+logx) + log(x^x)

=> dy/dx = y. ( x. (1+logx) + log(x^x))

=> ⭐dy/dx = x^x^x ( x. (1+logx) +log(x^x))

Answer 2

To Find:

The derivative of:

$y = (x {}^{x} ) {}^{x}$

Solution:

On Opening the parenthesis, we get:

$y = (x ){}^{x {}^{2} }$

Now,

$\frac{d}{dx} [x {}^{ {x}^{2} } ] = \frac{d}{dx} [(e {}^{ log_{e}(x) }) {}^{ {x}^{2} } ]$

$\frac{d}{dx} [x {}^{ {x}^{2} } ] = \frac{d}{dx} [(e ^{ln(x) \times x {}^{2} } ) ]$

Now, Let,

u=ln(x).x²

»y=e^u

Now, using the chain rule;

$\frac{dy}{dx} = \frac{d}{du} (y) \times \frac{d}{dx} (u)$

We get:

$\frac{dy}{dx} = \frac{d}{du} [e {}^{u} ] \times \frac{d}{dx} [ln(x). {x}^{2} ]$

Now, using the product rule, we get:

$= e {}^{u} \times ( \frac{1}{x} .x {}^{2} + ln(x).2x)$

$= e {}^{ln(x). {x}^{2} } .(x + ln(x).2x)$

$= (x) {}^{ {x}^{2} + 1 } +( 2x) {}^{ {x}^{2} + 1} . log_{e}(x)$

Therefore, the derivative of the given function is given by:

$\frac{dy}{dx} = (x) {}^{ {x}^{2} + 1 } +( 2x) {}^{ {x}^{2} + 1} . log_{e}(x)$