How do you differentiate $y = x^{\sqrt{x}}$ $y = x ^ sqrt(x)$ ?

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How do you differentiate $y = x^{\sqrt{x}}$ $y = x ^ sqrt(x)$ ?

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Answer 1 · 2016-10-05T19:08:45

$\frac{d y}{d x} = \frac{1}{2} x^{- \frac{1}{2} + \sqrt{x}} (2 + {log}_{e} (x))$ $dy/dx=1/2 x^(-1/2 + sqrt[x]) (2 + Log_e(x))$

Explanation:

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

Applying the $log$ $log$ transformation ti both sides

${log}_{e} y = \sqrt{x} {log}_{e} x$ $log_ey=sqrt(x) log_e x$ so

$\frac{d y}{y} = (\frac{1}{2} \frac{{log}_{e} x}{\sqrt{x}} + \frac{\sqrt{x}}{x}) d x$ $dy/y = (1/2log_e x/sqrt(x)+sqrt(x)/x)dx$ so

$\frac{d y}{d x} = (\frac{1}{2} \frac{{log}_{e} x}{\sqrt{x}} + \frac{\sqrt{x}}{x}) y = (\frac{1}{2} \frac{{log}_{e} x}{\sqrt{x}} + \frac{\sqrt{x}}{x}) x^{\sqrt{x}}$ $dy/dx = (1/2log_e x/sqrt(x)+sqrt(x)/x)y = (1/2log_e x/sqrt(x)+sqrt(x)/x)x^sqrt(x)$

Finally

$\frac{d y}{d x} = \frac{1}{2} x^{- \frac{1}{2} + \sqrt{x}} (2 + {log}_{e} (x))$ $dy/dx=1/2 x^(-1/2 + sqrt[x]) (2 + Log_e(x))$

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How do you differentiate $y = x^{\sqrt{x}}$ $y = x ^ sqrt(x)$ ?

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