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?
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?
$\displaystyle u = x^{\sqrt{x}}$
$\displaystyle \ln{u} = \sqrt{x} \cdot \ln{x}$
$\displaystyle \frac{u'}{u} = \frac{1}{\sqrt{x}} + \frac{\ln{x}}{2\sqrt{x}} = \frac{2+\ln{x}}{2\sqrt{x}}$
$\displaystyle u' = x^{\sqrt{x}} \cdot \frac{2+\ln{x}}{2\sqrt{x}}$
finally ...
$\displaystyle \frac{dy}{dx} = -\sin\left(x^{\sqrt{x}}\right) \cdot x^{\sqrt{x}} \cdot \frac{2+\ln{x}}{2\sqrt{x}}$