Question

if y=x^√x then dy/dx is ​

Answer 1

Answer:

Given , y=xx√

⇒logy=x−−√logx

⇒1ydydx=x−−√x+logx×12x−−√

⇒dydx=y(2x−−√+x−−√logx2x)=y2x−−√(2+logx)

Answer 2

Solution :

We have,

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

Taking log both sides

$\leadsto\sf\log y = \log x^{\sqrt{x}}$

$\blue{\leadsto\sf\log y = \sqrt{x}\log x}$

Now differentiating both sides and applying chain rule,

$\sf\implies\dfrac{1}{y}.\dfrac{dy}{dx} =\left( \dfrac{d}{dx}(\sqrt{x}).\log x + \sqrt{x}.\dfrac{d}{dx}(\log{x})\right)$

$\sf\implies \dfrac{1}{y}.\dfrac{dy}{dx} =\left( \dfrac{1}{2\sqrt{x}}.\log x + \sqrt{x}.\dfrac{1}{x}\right)$

$\sf \implies \dfrac{1}{y}.\dfrac{dy}{dx} =\left( \dfrac{\log{x}}{2\sqrt{x}} +\dfrac{1}{\sqrt{x}}\right)$

$\sf\implies \dfrac{1}{y}.\dfrac{dy}{dx} =\left( \dfrac{\log{x}}{2\sqrt{x}} +\dfrac{1}{\sqrt{x}}\right)$

$\sf \implies \dfrac{1}{y}.\dfrac{dy}{dx} =\dfrac{1}{\sqrt{x}}\left(\dfrac{\log{x}}{2} +1\right)$

$\sf \implies\dfrac{1}{y}.\dfrac{dy}{dx} =\dfrac{1}{\sqrt{x}}\left(\dfrac{\log{x}+2}{2} \right)$

$\sf\implies \dfrac{dy}{dx} =y\times\dfrac{1}{\sqrt{x}}\left(\dfrac{\log{x}+2}{2} \right)$

$\sf\implies\dfrac{dy}{dx} ={\bf x^{\sqrt{x}}}\times\dfrac{1}{\sqrt{x}}\left(\dfrac{\log{x}+2}{2} \right)$

$\sf \implies \pink{\dfrac{dy}{dx} ={\bf x^{\sqrt{x}}}\times\left(\dfrac{\log{x}+2}{2\sqrt{x}} \right)}$

Additional Information :-

$\boxed{\begin{array}{c|c} \tt f(x)& \tt f'(x)\\\underline{\qquad\qquad}&\underline{\qquad\qquad}\\ \\ \sf x^n&\sf nx^{n-1}\\\\\sf {\log_ea}&\sf{\dfrac{1}{x}}\\\\\sf {e^x}&\sf {e^x} \\ \\ \sf log_{a}x & \sf\dfrac{1}{x}\log_ea\\\\ \sf a^x&\sf a^x\log_ea\\\\\sf\sqrt{x}&\sf\dfrac{1}{2\sqrt{x}}\end{array}}$