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My problem is; find the derivative of $$y= \frac{\cos(\pi x)}{\sin(\pi x) + \cos(\pi x)}$$

Can someone please explain to me how to do the process in detail. I get the fact that you can use the quotient rule along with the chain rule, it's just that the $\sin(\pi x) + \cos(\pi x)$ throws me off when i try to get the inner derivative of it. I would appreciate if you can help. Thank you.

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Differentiation is "linear" meaning you can tackle sums one term at a time. So $\sin(\pi x)$ and $\cos(\pi x)$ can be differentiated separately and then put back together.

As for each of those terms, $\sin(\pi x)$ can be pulled apart as: $y=\sin(u)$ and $u=\pi x$. Then the chain rule says: $$\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx} = \cos(u) \cdot \pi = \pi \cos(\pi x)$$

Likewise for the cosine term. So... $$\frac{d}{dx}\left[\sin(\pi x)+\cos(\pi x)\right] = \pi \cos(\pi x) - \pi \sin(\pi x)$$

Putting this together with the quotient rule one finds that... $$\frac{d}{dx}\left[\frac{\cos(\pi x)}{\sin(\pi x)+\cos(\pi x)}\right]$$ $$= \frac{-\pi \sin(\pi x)\left(\sin(\pi x)+\cos(\pi x)\right)-\cos(\pi x)\left(\pi \cos(\pi x) - \pi \sin(\pi x)\right)}{\left(\sin(\pi x)+\cos(\pi x)\right)^2}$$ I'll leave it to you to simplify. :)

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Use $$y'=(\frac{u}{v})'=\frac{v\cdot u'-u\cdot v'}{v^2},$$ where $$u=\cos \pi x \Rightarrow u'=-\pi \sin \pi x $$ $$v=\sin \pi x+\cos \pi x \Rightarrow v'=\pi\cos \pi x-\pi \sin \pi x. $$ Then, $$y'=\frac{(\sin \pi x+\cos \pi x)\cdot (-\pi \sin \pi x)-(\cos \pi x)(\pi\cos \pi x-\pi\sin \pi x)}{(\sin \pi x+\cos \pi x)^2}= $$ $$=\frac{-\pi \sin^2\pi x-\pi \sin \pi x\cos \pi x-\pi \cos^2\pi x+\pi \sin \pi x\cos \pi x}{(\sin \pi x+\cos \pi x)^2}= $$ $$=\frac{-\pi(\sin^2\pi x +\cos^2\pi x)}{\sin^2\pi x+2\sin \pi x\cos \pi x+\cos^2\pi x}= $$ $$=\frac{-\pi}{1+2\sin \pi x\cos \pi x}=-\frac{\pi}{1+\sin 2\pi x}. $$

Ps.: Note that I use the equalities: $\sin^2 \alpha +\cos^2\alpha =1$ and $\sin 2\alpha=2\sin \alpha\cos \alpha$.

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