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Question Number 188037 by Michaelfaraday last updated on 25/Feb/23

find (dy/dx)  y=2x^(√x)

$${find}\:\frac{{dy}}{{dx}} \\ $$$${y}=\mathrm{2}{x}^{\sqrt{{x}}} \\ $$

Answered by horsebrand11 last updated on 25/Feb/23

 ln y=(√x) ln (2x)   (1/y).y′=((ln (2x))/(2(√x)))+((2(√x))/(2x))   y′=2x^(√x)  (((ln (2x)+2)/(2(√x))))   y′= (ln (2x)+2)x^((√x)−(1/2))

$$\:\mathrm{ln}\:{y}=\sqrt{{x}}\:\mathrm{ln}\:\left(\mathrm{2}{x}\right) \\ $$$$\:\frac{\mathrm{1}}{{y}}.{y}'=\frac{\mathrm{ln}\:\left(\mathrm{2}{x}\right)}{\mathrm{2}\sqrt{{x}}}+\frac{\mathrm{2}\sqrt{{x}}}{\mathrm{2}{x}} \\ $$$$\:{y}'=\mathrm{2}{x}^{\sqrt{{x}}} \:\left(\frac{\mathrm{ln}\:\left(\mathrm{2}{x}\right)+\mathrm{2}}{\mathrm{2}\sqrt{{x}}}\right) \\ $$$$\:{y}'=\:\left(\mathrm{ln}\:\left(\mathrm{2}{x}\right)+\mathrm{2}\right){x}^{\sqrt{{x}}−\frac{\mathrm{1}}{\mathrm{2}}} \\ $$$$ \\ $$

Commented by Michaelfaraday last updated on 25/Feb/23

thanks sir

$${thanks}\:{sir} \\ $$

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