Derivative of $\cos^x \theta$

Question

I wondered how to calculatethe following derivative : $$\frac{\mathrm d}{\mathrm dx}(\cos^x\theta)$$ where $x\in \mathbb{R}$.

It is kind of silly question, but I could not find the way to search at the internet.

Thanks.

Answer 1

Sounds simple, if I get the question rightly. $$\dfrac{d(\cos^x\theta )}{dx}=\cos^{x}\theta\ln\cos\theta$$

Answer 2

If $\theta$ is just a constant (meaning that $x$ and $\theta$ are independent variables), then :

$$\cos^x\theta=(\cos \theta)^x = e^{x \ln (\cos \theta)}$$

and thus $(\cos^x \theta)' = (e^{x \ln(\cos \theta)})' = (x\ln(\cos \theta))' \cdot e^{x \ln (\cos \theta)} = \ln(\cos \theta) e^{x \ln(\cos \theta)}= \ln(\cos \theta)\cos^x \theta.$

Answer 3

As noted: $$y=\cos^x\theta=(\cos\theta)^x=e^{x\ln(\cos\theta)}$$

Treating $\theta$ as another variable: $$y'=(x\ln(\cos\theta))'\cdot e^{x\ln(\cos\theta)}$$

Let $p=x\to p'=1$ and $q=\ln(\cos\theta)\to q'=\frac{-\sin\theta}{\cos\theta}=-\tan\theta\frac{d\theta}{dx}$

The product rule: $$(pq)'=p'q+q'p$$ Tells us: $$(x\ln(\cos\theta))'=\ln(\cos\theta)-x\tan\theta\frac{d\theta}{dx}$$ Subbing this back in, noting $e^{x\ln(\cos\theta)}=\cos^x\theta$, we get: $$y'=\cos^x\theta\ln(\cos\theta)-x\cos^x\theta\tan\theta\frac{d\theta}{dx}$$