Question

find the dy/dx if y=x sqrt{lnx} ​

Answer 1

I have given the attachment

You will understand the problem when you know the formula.

Answer 2

Question :

Find $\dfrac{dy}{dx}$ if $y = x \sqrt{ \ logx}$

Theory :

${\red{\boxed{\large{\bold{Chain\:Rule}}}}}$

Let y=f(t) ,t = g(u) and u =m(x) ,then

$\dfrac{dy}{dx} = \dfrac{dy}{dt} \times \dfrac{dt}{du} \times \dfrac{du}{dx}$

Solution :

$y = x \sqrt{ \ logx}$

Differentiate with respect to x

$\dfrac{dy}{dx} = \sqrt{ \log x} \dfrac{d(x)}{dx} + x \dfrac{d( \sqrt{ \log x} )}{dx}$

$\implies \dfrac{dy}{dx} = \sqrt{ \log x} \times 1 + x \times \dfrac{1}{2 \sqrt{ \log x} } \times \dfrac{1}{x}$

$\implies \dfrac{dy}{dx} = \sqrt{ \log x} + \dfrac{1}{2 \sqrt{ \log x} }$

$\implies \dfrac{dy}{dx} = \dfrac{2 \log x + 1}{2 \sqrt{ \log x} }$

which is the required solution!

_________________________

Some formulas related to Differention:

$1) \dfrac{d(sinx)}{dx} = cosx$

$2) \dfrac{d(x {}^{n}) }{dx} = nx {}^{n - 1}$

$3) \dfrac{d(constant)}{dx} = 0$