Need derivative of y = cos (x^sqrt(x))

**potatoes** · August 3rd 2009, 04:22 PM

I'm trying do it using log differentiation but I'm not really sure how to use log rules to bring down the sqrt (x) . Can I do ln y = sqrt(x) ln cos x ? It seems like I'm wrongly taking the sqrt (x) out of the bracket if I do that?

**skeeter** · August 3rd 2009, 04:28 PM

Originally Posted by potatoes

I'm trying do it using log differentiation but I'm not really sure how to use log rules to bring down the sqrt (x) . Can I do ln y = sqrt(x) ln cos x ? It seems like I'm wrongly taking the sqrt (x) out of the bracket if I do that?

do this ...

$y = \cos(u)$

$\frac{dy}{dx} = -sin(u) \cdot \frac{du}{dx}$

let $u = x^{\sqrt{x}}$ and use logarithmic differentiation to find $\frac{du}{dx}$

**potatoes** · August 3rd 2009, 04:41 PM

Ok thanks.

So that gives me dy/dx = -sin u * (sqrt(x)lnx + 2sqrt(x)/2x) * u

At this point does the derivative become dy/du? and then I can just leave it like that? or do I sub in x^sqrt(x) for u?

**skeeter** · August 3rd 2009, 04:50 PM

$u = x^{\sqrt{x}}$

$\ln{u} = \sqrt{x} \cdot \ln{x}$

$\frac{u'}{u} = \frac{1}{\sqrt{x}} + \frac{\ln{x}}{2\sqrt{x}} = \frac{2+\ln{x}}{2\sqrt{x}}$

$u' = x^{\sqrt{x}} \cdot \frac{2+\ln{x}}{2\sqrt{x}}$

finally ...

$\frac{dy}{dx} = -\sin\left(x^{\sqrt{x}}\right) \cdot x^{\sqrt{x}} \cdot \frac{2+\ln{x}}{2\sqrt{x}}$

**potatoes** · August 3rd 2009, 04:52 PM

Ok great. Thanks so much.

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