Question

Differentiation of y= x raise to root x


Please don't give foul answers​

Answer 1

Step-by-step explanation:

y = x^√x

Taking log

logy = √xlogx

D . w .r .to .x

1/y * dy/dx = √x (1/x) + logx * 1/2√x

1/y * dy/dx = 1/√x + logx / 2√

1/y * dy/dx = (2+logx)/2√x

dy/dx = (2y+logx^y)/2√x

Answer 2

Answer:

$\large\boxed{\sf{{ {x} }^{ (\sqrt{x}-\frac{1}{2}) } (1 + \frac{ logx }{2} )}}$

Step-by-step explanation:

$y = {x}^{ \sqrt{x} }$

Taking log both sides, we get

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

Concept Map :-

• $\dfrac{d}{dx} log(x) = \dfrac{1}{x}$

• $\dfrac{d}{dx} uv = u \dfrac{dv}{dx} + v \dfrac{du}{dx}$

• $\dfrac{d}{dx} {x}^{n} = n {x}^{n - 1}$

Now, differentiate both sides wrt x, we get

$= > \dfrac{1}{y} \dfrac{dy}{dx} = \sqrt{x} \dfrac{d}{dx} log(x) + log(x) \dfrac{d}{dx} \sqrt{x} \\ \\ = > \frac{1}{y} \frac{dy}{dx} = \sqrt{x} \times \frac{1}{x} + log(x) \times \frac{1}{2 \sqrt{x} } \\ \\ = > \frac{1}{y} \frac{dy}{dx} = \frac{1}{ \sqrt{x} } + \frac{ log(x) }{2 \sqrt{x} } \\ \\ = > \frac{dy}{dx} = y( \frac{1}{ \sqrt{x} } + \frac{ log(x) }{2 \sqrt{x} } ) \\ \\ = > \frac{dy}{dx} = {x}^{ \sqrt{x} } ( \frac{1}{ \sqrt{x} } + \frac{ log(x) }{2 \sqrt{x} } ) \\ \\ = > \frac{dy}{dx} = { x}^{ (\sqrt{x}-\frac{1}{2}) } (1 + \frac{ logx }{2} )$