derivativ

**hoger** · December 7th 2008, 01:37 AM

find f prim (1) when f(x)= x^1+sqrt x .

thanks

**running-gag** · December 7th 2008, 01:49 AM

Hi

$f(x) = x^{1+\sqrt{x}} = e^{(1+\sqrt{x})ln(x)}$

The first derivative of

$e^{u(x)}$ is $u'(x) e^{u(x)}$

Here $u(x) = (1+\sqrt{x})ln(x)$

u(x) is a product of 2 functions
$u'(x) = \frac{1}{2 \sqrt{x}}ln(x) + \frac{1+\sqrt{x}}{x}$

**hoger** · December 7th 2008, 03:45 AM

hi !
I dont know this is power fuction or exponetial function can you explain more .
thank you

**skeeter** · December 7th 2008, 03:54 AM

familiar with logarithmic differentiation?

$y = x^{1+\sqrt{x}}$

$\ln{y} = \ln\left(x^{1+\sqrt{x}}\right)$

$\ln{y} = (1 + \sqrt{x})\ln{x}<br />$

$\frac{d}{dx}[\ln{y} = (1 + \sqrt{x})\ln{x}]$

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

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

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

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