Nth Order Derivative of f(x): Sin^4(x)+Cos^4(x) & x^n/(1-x)

[shrody]

Homework Statement

The exercise goes "Determine d^n*f/dx^n for the 2 exercises:"
a) f(x)=sin^4(x) + cos^4(x)
b) f(x)= x^n/(1-x)

Homework Equations

The Attempt at a Solution

The only idea i had was for the second example, where i think its right to use a rule from Leibniz but I'm not sure...

 

[Mentallic]

I've never done any of these before, so please forgive me if I'm way off and not being helpful at all.


a) couldn't you express $sin^4x$ and $cos^4x$ in terms of multiple angles without powers, and then take the derivatives of those?


b) you can make the fraction like this: $$\frac{-(1-x^n)+1}{1-x}$$

and then they split up as so: $$\frac{-(1-x^n)}{1-x}+(1-x)^{-1}$$

The first fraction can be... expanded (for lack of a better word) into $-(1+x+x^2+...+x^{n-1})$ and I'm sure it's clear what to do with the second fraction.

 

[shrody]

i don't know about the first one with the angles, i think the whole point of the exercise is to keep the sin and cos and the second one kinda confused me...

 

[Mentallic]

Well you would still keep the sin and cos, just that theyre expressed as

$$sin^nx=Asin(nx)+Bsin((n-1)x)...$$

I know you could find the relationship easily for the n-th derivative from that, but if that isn't allowed, then maybe this will help?

$$y=sin^4x$$

$$\frac{dy}{dx}=4sin^3xcosx$$

$$\frac{d^2y}{dx^2}=-16sin^4x+12sin^2x$$

Notice how the $sin^4x$ appears again. Maybe it's something, probably it's not, but I'm just putting that out there in case it helps.


For the second, is that "confused me" or "still confuses me"?

$$(x^n-1)=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$$

dividing both sides by $x-1$ will show you how to get the long expansion thing.

Was this the problem?