# Solve 1

• Mar 27th 2006, 05:59 PM
Hallah_az
Solve 1
How would I solve these problems using a short cut formula, simplify results, collecting like terms if possible. I am totally lost with derivative, quotient and product rule.

f(x) =-8

g(x)=4x^-5+20x^-1

k(x)=15x 2/3

r(x)=16x^7-8x^5+4x^3-2x+1

f(x) = (√x+1)(Vx-1)
• Mar 27th 2006, 06:35 PM
Jameson
Ok. Some basic rules. The derivative of a constant is zero. That could be 1,2,3, pi, e, phi, etc. - the derivatives of all of these are zero.

If you have a polynomial in the form of $y=x^n$, then $y'=nx^{n-1}$ So you can apply both of these rules to your f(x),g(x), and r(x).

On k(x), is that supposed to be $15x^{\frac{2}{3}}$, or $15x*\frac{2}{3}$? I can't tell. And on the second f(x) term, is that supposed to be $\frac{\sqrt{x+1}}{\sqrt{x-1}}$?
• Mar 27th 2006, 06:38 PM
ThePerfectHacker
Quote:

Originally Posted by Hallah_az
How would I solve these problems using a short cut formula, simplify results, collecting like terms if possible. I am totally lost with derivative, quotient and product rule.

f(x) =-8

g(x)=4x^-5+20x^-1

k(x)=15x 2/3

r(x)=16x^7-8x^5+4x^3-2x+1

f(x) = (√x+1)(Vx-1)

$f'(x)=(-8)'=0$

$g'(x)=(4x^{-5}+20x^{-1})'=-20x^{-6}-20x^{-2}$

$k'(x)=(15x^{2/3})'=15\cdot\frac{2}{3}x^{-1/3}=10x^{-1/3}$

$r'(x)=(16x^7-8x^5+4x^3-2x+1)'=$ $112x^6-40x^4+12x^2-2$

$f'(x)=[(\sqrt{x}+1)(\sqrt{x}-1)]'=[x-1]'=1$

Next time place the thread in the appropraite section.