What is $\frac{d\left( (\cos(x))^{\cos(x)}\right)}{dx}$?

Question

How would you work something like this out?

Are there similar problems to $$\frac{d\left( (\cos(x))^{\cos(x)}\right)}{dx}$$ which could be worked out the same way?

Answer 1

Hint: Given proper domain for the function so that $\cos(x) >0$ we can write:
$$(\cos x)^{\cos x} = e^{(\cos x) \ln(\cos x)}$$

Answer 2

The go-to way for the derivative of $f^g$ is $(f(x)^{g(x)})'=(e^{g(x)\ln f(x)})'$ and then use chain rule. Or knowing by heart the formula that follows directly from this procedure. Of course, when $f(x)>0$, because for some reason we appear to be in a moment where people seem to be very adamant about saying stuff like "$\left(\frac1x\right)'=-\frac{1}{x^2}$ holds only if $x\ne 0$".

Answer 3

A generalized power rule is $$(f^g)'=gf^{g-1}\cdot f' + f^g\log f \cdot g'$$

This generalizes the power rule and the exponential rule simultaneously. It is easily obtained by using implicit differentiation. But it is extremely easy to remember if you look at the components of the sum: when $g$ is constant, the factor $g'$ vanishes and you recover the familiar power rule for constant exponents; when $f$ is constant, the factor $f'$ vanishes and you recover the familiar exponential rule for constant bases. Knowing those simple rules, you can piece together the rule easily without rederiving it.

In your case, $f(x)=g(x)=\cos x$, so the result is $$((\cos x)^{\cos x})'=\cos x(\cos x)^{\cos x - 1}(-\sin x) + (\cos x)^{\cos x}\log\cos x(-\sin x)$$ $$=\boxed{-(\cos x)^{\cos x}\cdot \sin x\cdot(1 + \log\cos x)}$$

Implicitly, this is defined where everything makes sense ($\cos x > 0$, e.g.).

Answer 4

For any function of the type $u(x) ^{v(x)} $, use the logarithmic rule

ie, let $y =u^v$

$\implies ln(y) = vln(u) $

Differentiate wrt x

$\frac{dy}{dx} \frac 1y = v'ln(u) +\frac vu u'$

$\frac{dy}{dx} =y\big(v'ln(u) +\frac vu u'\big) $

For $ cos(x)^{cos(x)} $

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

Answer 5

So $\cos(x)^{\cos(x)}=e^{\cos(x)ln(\cos(x))}$ and $$\frac{d}{dx} e^{\cos(x)ln(\cos(x))}=e^{\cos(x)ln(\cos(x)} \cdot (-\sin(x) ln(\cos(x))+ -\sin(x)=\cos(x)^{\cos(x)} \cdot -\sin(x) ln(\cos(x))-\sin(x)$$