How do you differentiate #f(x) =(1-x)/(x^3-6)# using the quotient rule?

Question

Emilia Aguilar · Accepted Answer

#d/dx (f(x)) = (2x^3-3x^2+6)/((x^3 - 6)^2)#

First, let's clarify what the quotient rule is:

#d/dx (f(x)/(g(x))) = (g(x)f'(x) - f(x)g'(x))/(g(x))^2#

Or to make remembering the rule simpler, let #f(x)# be "#high#," being the upper function and let #g(x)# be "#low#," being the lower function to give us:

#d/dx ((high)/(low)) = (low*d'(high) - (high*d'(low)))/(low)^2#

where #d'# denotes the "derivative of."

The formula has a nice flow for memorization and can be read as "low d high minus high d low over low squared."

In the case #f(x) = (1-x)/(x^3 - 6)#, let "#low#" denote #x^3 - 6# and "#high#" denote #1-x#

Using the rule of quotient,

#d/dx ((high)/(low)) = ((x^3 - 6)*d'(1-x) - ((1-x)*d'(x^3 - 6)))/(x^3 - 6)^2#

Now switching #((high)/(low))# to #f(x)# for simplicity,

#d/dx (f(x)) = ((x^3 - 6)⋅(-1) - ((1-x)*(3x^2)))/(x^3 - 6)^2#

Knowing that by the power rule #d/dx (x^n) = nx^(n-1)#

And that the derivative of a constant is #0#.

#d/dx (f(x)) = ((-x^3 + 6) - ((3x^2-3x^3)))/(x^3 - 6)^2#

#d/dx (f(x)) = (-x^3 + 6 - 3x^2+3x^3)/(x^3 - 6)^2 = (2x^3-3x^2+6)/((x^3 - 6)^2)#

Emilia Aguilar · Answer

undefined To differentiate $ f(x) = \frac{{1-x}}{{x^3 - 6}} $ using the quotient rule, follow these steps:

1. Identify $ u $ and $ v $:
   $ u = 1 - x $
   $ v = x^3 - 6 $

2. Find $ u' $ and $ v' $ (the derivatives of $ u $ and $ v $):
   $ u' = -1 $
   $ v' = 3x^2 $

3. Apply the quotient rule:
   $ f'(x) = \frac{{v \cdot u' - u \cdot v'}}{{v^2}} $

4. Substitute the values:
   $ f'(x) = \frac{{(x^3 - 6)(-1) - (1 - x)(3x^2)}}{{(x^3 - 6)^2}} $

5. Simplify the expression to get the derivative.