Question

If y= (√x)^ √ x^ √x^ √x find its derivative.

Answer 1

Answer :

$\boxed{\sf \frac{dy}{dx} = ( { \sqrt{x}) }^{(x \sqrt{x} )} ( \frac{ \sqrt{x} }{2} + log \sqrt{x}( \frac{3}{2} \sqrt{x} )) }$

Step-by-step explanation:

let the function is,

$\sf y = ( { \sqrt{x} })^{( { \sqrt{x}) }^{ ({ \sqrt{x}) }^{ (\sqrt{x} )} } }$

$\sf y = {( \sqrt{x}) }^{( { \sqrt{x} )}^{( \sqrt{x} \times \sqrt{x} )} }$

$\sf y = {( \sqrt{x}) }^{(x \sqrt{x} )}$

taking log on both the sides

$\sf log(y) = log ({ \sqrt{x} }^{(x \sqrt{x}) } )$

we know, log rule,

$\sf log m ^n = n log m$

therefore,

$\sf log(y) = (x \sqrt{x}) log ({ \sqrt{x} })$

Differentiation wrt x

$\sf \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} ( (x \sqrt{x}) log ({ \sqrt{x} }))$

we know, product rule for differentiation

[if y = u. v, then dy /dx = u.(dv/dx)+v. (du/dx)]

therefore,

$\sf \frac{1}{y} \times \frac{dy}{dx} = x \sqrt{x} \times \frac{1}{ \sqrt{x} } \times \frac{1}{2 \sqrt{x} } + log \sqrt{x} (x \times \frac{1}{2 \sqrt{x} } + \sqrt{x} )$

$\sf = \frac{x}{2 \sqrt{x} } + log \sqrt{x} ( \frac{x + 2x}{2 \sqrt{x} } )$

$\sf = \frac{x}{2 \sqrt{x} } + log \sqrt{x} ( \frac{3x}{2 \sqrt{x} })$

$\sf = \frac{ \sqrt{x} \times \sqrt{x} }{2 \sqrt{x} } + log \sqrt{x} ( \frac{3 \sqrt{x} \times \sqrt{x} }{2 \sqrt{x} } )$

$\sf = \frac{ \sqrt{x} }{2} + log \sqrt{x} (\frac{3}{2} \sqrt{x} )$

$\sf \frac{dy}{dx} = y( \frac{ \sqrt{x} }{2} + log \sqrt{x} ( \frac{3}{2} \sqrt{x} ))$

Now, put the value of y

$\boxed{ \sf \frac{dy}{dx} = ( { \sqrt{x}) }^{(x \sqrt{x} )} ( \frac{ \sqrt{x} }{2} + log \sqrt{x}( \frac{3}{2} \sqrt{x} )) }$