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

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

Answered by horsebrand11 last updated on 25/Feb/23

$$\ln y=\sqrt{x}\ln \left(2x\right)$$$$\frac{1}{y}.y^{\prime }=\frac{\ln \left(2x\right)}{2\sqrt{x}}+\frac{2\sqrt{x}}{2x}$$$$y^{\prime }=2x^{\sqrt{x}}\left(\frac{\ln \left(2x\right)+2}{2\sqrt{x}}\right)$$$$y^{\prime }=(\ln \left(2x\right)+2)x^{\sqrt{x}-\frac{1}{2}}$$$$$$

Commented by Michaelfaraday last updated on 25/Feb/23

$$thankssir$$

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