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if $y = x^{(x+1)^\frac12}$

then how can I get the first derivative of $y$?

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Write $y=\exp\bigl((x+1)^{1/2}\ln x\bigr)$ and use the chain rule. – David Mitra Feb 22 '13 at 2:45
Do you mean $(x^{(x+1)})^{(1/2)}$ or $x^{((x+1)^{1/2})}$? – Ben Feb 22 '13 at 2:57
@Ben I mean your second choice. – DUKE Feb 22 '13 at 3:38
up vote 1 down vote accepted

Another way to solve it is to take $\ln$ of both sides, and apply implicit differentiation:

$$y=x^{(x+1)^{\frac {1}{2}}}$$

$$\ln y=\ln x^{(x+1)^{\frac {1}{2}}}$$

Rewriting using $\log$ properties:

$$\ln y=(x+1)^{\frac {1}{2}} \cdot \ln x$$

Now take the derivative implicitly:

$$\frac{1}{y}\cdot y' = \frac{1}{2}(x+1)^{-\frac{1}{2}} \ln x + (x+1)^{\frac {1}{2}} \cdot \frac{1}{x}$$

We want $y'$, so we multiply $y$ on both sides to isolate $y'$.

$$ y' = y \cdot \left( \frac{1}{2}(x+1)^{-\frac{1}{2}} \ln x + (x+1)^{\frac {1}{2}} \cdot \frac{1}{x} \right)$$

We want the right hand side to be in terms of $x$. The original problem says $y=x^{(x+1)^{\frac {1}{2}}}$, so

$$ y' = \big( x^{(x+1)^{\frac {1}{2}}} \big) \cdot \left( \frac{1}{2}(x+1)^{-\frac{1}{2}} \ln x + (x+1)^{\frac {1}{2}} \cdot \frac{1}{x} \right)$$

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$$y=x^{(x+1)^{1/2}}=x^{\sqrt{x+1}}=e^{\sqrt{x+1}\log x}$$

and since

$$(\sqrt{x+1}\log x)'=\frac{\log x}{2\sqrt{x+1}}+\frac{\sqrt{x+1}}{x}=\frac{x\log x+2x+2}{2x\sqrt{x+1}}$$

we get, applying as suggested the chaing rule, that

$$y'=y\frac{x\log x+2x+2}{2x\sqrt{x+1}}=x^{\sqrt{x+1}}\cdot\frac{x\log x+2x+2}{2x\sqrt{x+1}}$$

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In general, if $y = x^{f(x)}$ then $y = (e^{\ln(x)})^{f(x)} = e^{f(x)\ln(x)}$

Now the derivative should be easy: $y' = e^{f(x)\ln(x)}\cdot \frac{d}{dx}(f(x)\ln(x))$, by the chain rule.

This simplifies. The first part turns back to the original expression: $y'=x^{f(x)}\cdot \frac{d}{dx} (f(x)\ln(x))$

Or, in other words, $y' = y\cdot(f'(x) \ln(x) + f(x)/x)$.

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