Differentiate the following function (Fraction)?

Question

g(x) = x^2 +7 / x^3 -2

Answer 1

You didn't write your problem clearly. I don't know if g(x) is:

1) x² + (7/x³) - 2, or
2) (x² + 7) / (x³ - 2)

The quality of the answers you receive are in direct relation to the quality of the question you write! :-)

Assuming you meant (1):

g '(x) = 2x - 21x^-4

Assuming you meant (2):

g '(x) = (-x^4 - 21x² - 4x) / (x³ - 2)²

EDITED TO ADD:

Hi Jeff, I got your email but couldn't respond; yahoo said your email was "unconfirmed" whatever that means! So, I am responding to you here. Hope you see this!

YOU WROTE: "I was wondering if you could explain how to work through this problem"

MY RESPONSE:

Hi Jeff, sure I am happy to explain. Basically, when you differentiate a fraction f(x) over g(x) the rule is "bottom times derivative of top, minus top times derivative of bottom, all over bottom squared" -- i.e.,

[ (g(x)*f '(x)) - (f(x)*g '(x)) ] / (g(x))^2

So in your problem f(x) = x^2 - 7
g(x) = x^3 - 2

So derivative = [ (x^3 - 2)*(2x) - (x^2 - 7)*3x^2 ] / (x^3 - 2)^2

Then just multiply the numerator and combine like terms to get:

[ -x^4 - 21x^2 - 4x ] / (x^3 - 2)^2

Did that help? Let me know if you have additional questions!

.

Answer 2

The derivative of a fraction is

the derivative of the numerator times the denominator
minus
the derivative of the denominator times the numerator
all divded by
the square of the denominator

[2x * (x^3 -2) - (3x^2) * (x^2 + 7)] / (x^3 - 2)^2

(-x^4-21x^2-4x)/(x^3-2)^2

Answer 3

to get the derivartive of a fraction you should use the following rule:
Denominator * derivative of numerator minus numerator times derivative of denominator and divide all of that by the square of the denominator.

dy/dx=[(x^3-2)(2x) -(x^2+7)3x^2)](X^3-2)^2
=(2x^4-4x -3x^4 + 21x^2)/(x^3-2)^2
=(-x^4+21x^2 -4x)/(x^3-2)^2

Answer 4

f'(x)=(x^3-2)(2x)-(x^2+7)(3x^2)/(x^3-2)^2
=2x^4-4x-3x^4-21x^2/(x^3-2)2
=-x^4-21x^2-4x/(x^3-2)^