Question

If $y=y^{\sqrt{x}}+x^{\sqrt{e}}$, find $\frac{dy}{dx}$

Answer 1

$y=y^{\sqrt{x}}+x^{\sqrt{e}}$ if $y^{\sqrt{x}}=\mu$

$\Rightarrow y=\mu +x^{\sqrt{x}}$

$\frac{dy}{dx}=\frac{d\mu }{dx}=\sqrt{e}\left(x^{\sqrt{e}-1}\right)---\left(1\right)$

$y^{\sqrt{x}}=\mu \Rightarrow \sqrt{x}\log y=\log u$

$\Rightarrow \sqrt{x}+\frac{1}{y}\frac{dy}{dx}+\frac{1}{2\sqrt{x}}\log y=\frac{1}{\mu }\frac{d\mu }{dx}$

$\Rightarrow \frac{d\mu }{dx}=\mu (\frac{\sqrt{x}}{y}\frac{dy}{dx}+\frac{1}{2\sqrt{x}}\log y)----\left(2\right)$

$\therefore \frac{dy}{dx}=\frac{\mu \sqrt{x}}{y}\frac{dy}{dx}+\frac{\mu }{2\sqrt{x}}\log y+\sqrt{e}\left(x^{\sqrt{e}-1}\right)$

$\frac{dy}{dx}\left(\frac{y-\mu \sqrt{x}}{y}\right)=\frac{\mu }{2\sqrt{x}}\log y+\sqrt{e}\left(x^{\sqrt{e}-1}\right)$

$\Rightarrow \frac{dy}{dx}=\left(\frac{y}{y-\mu \sqrt{x}}\right)(\frac{\mu }{2\sqrt{x}}\log y+\sqrt{e}(x^{\sqrt{e}-1}\left)\right)$

where $\mu =y^{\sqrt{x}}$ Answer