Question

help pls im stuck
differentiate y=cospix/(sinpix+cospix)

Answer 1

what is meant by cospix and sinpix?

Answer 2

I think he meant: \(\cos(\pi x)\) and \(\sin(\pi x)\). If so, could I try to differentiate that expression?

\(\large{f(x) = \frac{\cos(\pi x)}{\sin(\pi x) + \cos(\pi x)}}\)

Let write it is this way:

\(z = \cos(y)g(y)\)

\(\large{g(y) = \frac{1}{\sin(y) + \cos(y)}}\)

\(\large{y = \pi x}\)

By applying the product rule:

\(\frac{d}{dx}z = -\sin(y)g(y) + g’(y)\cos(y)\)

Now we need to find \(g'(y)\).By applying the quotient rule (there is a chain as well):

\(\large{g’(y) = \frac{-(1)(y’\sin(y) + y’\cos(y))}{\sin^2(y)+2\sin(y)\cos(y)+\cos^2(y)}}\)

\(\large{g’(y) = \frac{-y’(\sin(y)+\cos(y))}{1+\sin(2y)}}\) <-- Applying trig identities in the denominator.

Finally:

\(\large{\frac{d}{dx}z = \frac{-\sin(y)}{\sin(y)+\cos(y)} +\frac{ -y’[\sin(y)+\cos(y)]\cos(y)}{ 1+\sin(2y)}}\)

\(\large{\frac{d}{dx}z = \frac{-\sin(\pi x)}{\sin(\pi x)+\cos(\pi x)} +\frac{ -\pi[\sin(\pi x)+\cos(\pi x)]\cos(\pi x)}{ 1+\sin(2\pi x)}}\)


I tried to find the least common denominator in order to have just one term, but it'll end up being a much messier expression than this one.

If someone can point out some mistake I’ve eventually made it would be great.

Answer 3

By the quotient rule, the answer is \[-\pi/(\cos \pi x + \sin \pi x)^2\] and I'm pretty sure that is correct. If you wish I can post all the steps.

Answer 4

(New version because old version got lost in equation mark-up)
\[y=\frac{\cos(\pi x)}{\sin(\pi x)+\cos(\pi x)}\]
Let:
\[u=\cos(\pi x), u'=-\pi\sin(\pi x)\]
and:
\[v=\sin(\pi x)+\cos(\pi x), v'= \pi\cos(\pi x)-\pi\sin(\pi x)\]
Using the quotient rule:
\[\frac{d}{dx}y = \frac{(-\pi\sin(\pi x))(\sin(\pi x)+\cos(\pi x))-(\cos(\pi x)(\pi\cos(\pi x)-\pi\sin(\pi x))}{(\sin(\pi x)+\cos(\pi x))^2}\]
Expand the numerator:
\[-\pi\sin^2(\pi x)-\pi\sin(\pi x)\cos(\pi x)-\pi\cos^2(\pi x)+\pi\sin(\pi x)\cos(\pi x)\]
After a simple cancellation and factoring out negative pi:
\[-\pi(\sin^2(\pi x)+\cos^2(\pi x))\]
If you remember your trignometry identities or do some quick pythagoras, you'll know that:
\[sin^2\phi+\cos^2\phi=1\]
So the whole numerator becomes negative pi and you're left with:
\[\frac{d}{dx}y=-\frac{\pi}{(\sin(\pi x)+\cos(\pi x))^2}\]
Which is what @kimberstout said.