What is the derivative of #(ln x) ^ cos x#?

Question

Christopher Agresta · Accepted Answer

#(lnx)^cosx ( -sinx ln lnx + cosx /(x lnx))#

Let #y=(ln x)^cos x#, so that ln y= cos x ln ln x

Now differentiate both sides,

#1/y dy/dx = -sinx ln ln x + cos x 1/lnx 1/x#

#dy/dx= (lnx)^cosx ( -sinx ln lnx + cosx /(x lnx))#

Ezekiel Alexander · Answer

undefined To find the derivative of $ (\ln x)^{\cos x} $, we use the chain rule:

Let $ u = \ln x $, and $ v = \cos x $. Then we have $ y = u^v $.

$ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} + \frac{dy}{dv} \cdot \frac{dv}{dx} $

$ \frac{dy}{du} = vu^{v-1} $
$ \frac{du}{dx} = \frac{1}{x} $
$ \frac{dy}{dv} = u^v \ln u $
$ \frac{dv}{dx} = -\sin x $

Plugging these into the chain rule:

$ \frac{dy}{dx} = v(\ln x)^{\cos x - 1} \cdot \frac{1}{x} + (\ln x)^{\cos x} \ln(\ln x) \cdot (-\sin x) $

$ \frac{dy}{dx} = \frac{\cos x \cdot (\ln x)^{\cos x - 1}}{x} - (\ln x)^{\cos x} \sin x \ln(\ln x) $

Therefore, the derivative of $ (\ln x)^{\cos x} $ is $ \frac{\cos x \cdot (\ln x)^{\cos x - 1}}{x} - (\ln x)^{\cos x} \sin x \ln(\ln x) $.