DERIVATIVES HOMEWORK!! please help!!?

Question

how do you take derivatives of:

f(x)=(4x^3 - 3x^2)/(5^7+1) ??

and:

f(x)=(8*(x^4 - 4x^2)^(1/2))

Please show all the steps!
Thank you so much!

HEY, WHOEVER ANSWERS THIS!!!
THE DENOMINATOR OF THE SECOND ONE IS :

(5x^7+1)

Answer 1

1st- you have to use the quotient rule. Which is the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator. DID YOU MEAN 5X^7
It will equal out to be ((5x^7)(12x^2-6x)-(4x^3-3x^2)(35x^6+1))/ (5x^7+1)^2. FOR SOME REASON YAHOO WONT SHOW THE WHOLE ANSWER.

Answer 2

1. 5^7 + 1 is a constant, so just use power rule: f'(x) = (12x^2 - 6x)/(5^7 + 1)

2. Apply the chain rule. f'(x) = 4(x^4 - 4x^2)^(-1/2)*(4x^3 - 8x)

Answer 3

1) y = f(X) = (4X^3 - 3X^2)/5^8

y ' = ?

= ((4X^3 - 3X^2)' - (5^8) (4X^3 - 3X^2))/(5^8)^2

= ((12X^2 - 6X) (5^8) - (0))/(5^16)

= ((5^8)(12X^2 - 6X))/(5^16)


2) y = f(X) = (8) (X^4 - 4X^2)^1/2

y = ?

= (8)' (X^4 - 4X^2)^1/2 + (X^4 - 4X^2)^1/2 (8)

= 0 + 1/2(X^4 - 4X^2)^-1/2 (4X^3 -8X) (8)

= 4(X^4 -4X^2)^-1/2 (4X^3 -8X)

Answer 4

if f(x) = g(x)/h(x)
then f' = (h(g') - g(h'))/(h^2)
where ' means derivative
g= 4x^3-3x^2 so g' = 12x^2-6x
h = 5x^7+1 so h' = 35x^6
...

if f(x) = (g(x))^(1/2),
f' = (1/2)((g)^(-1/2))g'

if g(x) = x^4 - 4x^2
f' = (1/2)
((x^4-4x^2)(-1/2))(4x^3-8x)

if h(x) = af(x)
h' = af' where a is constant for example 8

so f' = 8(1/2)
((x^4-4x^2)(-1/2))(4x^3-8x)
= 16x(x^2-2)/sqrt(x^2(x^2-4))
= 16(x^2-2)/sqrt(x^2-4)

Answer 5

y = [4x^3 - 3x^2] / [5x^7+1]
f(x) = top g(x) = bottom
[f(x)'*g(x) - f(x)*g(x)'] / [g(x)]^2 TOO LONG TO DO HERE! I'll try anyway.
[12x^2-6x]*[5x^7+1] - [4x^3 - 3x^2]*[35x^6] / [25x^14 + 10x^7 + 1]
do foil on both and you should get:
[60x^14+12x^2-30x^8-6x]-[140x^9-105x^8] / [25x^14 + 10x^7 + 1]
[60x^14-140x^9-135x^8+12x^2-6x] / [25x^14 + 10x^7 + 1]
NEXT
y = (8*(x^4 - 4x^2)^(1/2))
y = (8x^4 - 32x^2) ^ 0.5
y' = 0.5 [8x^4 - 32x^2] ^ -0.5 * [8x^4 - 32x^2]'
y' = 0.5 [8x^4 - 32x^2] ^ -0.5 * [32x^3 - 64x]
y' = [32x^3 - 64x] / [2 * [8x^4 - 32x^2] ^ 0.5]

Answer 6

if y=a*x^n, then y’=a*n*x^(n-1). If y=g(x)+h(x), then y’=g’+h’.
if y=g(h(x)), then y’=(dg/dh)*h’(x)
Now problem#1:
y=f(x)=(4/(5^7+1))*x^3-(3/(5^7+1))*x^2
y’=(4/(5^7+1))*3*x^2-(3/(5^7+1))*2*x=(12/(5^7+1))*x^2-(6/(5^7+1))*x=
=(6/(5^7+1))*x*(2x-1).
now problem#2:
y=8*sqrt(x^4-4*x^2),
y’=8*(1/2)*(x^4-4x^2)^(-1/2)*(x^4-4x^2)’=
=-(2/sqrt(x^4-4x^2))*(4*x^3-4*2*x)=-8*x*(x^2-2)/(x*sqrt(x^2-4))=-8(x^2-2)/sqrt(x^2-4)