How do you find the derivative of #y=3/(2x)^3#?

1 Answer
Nov 29, 2017

#-9/8x^-4 or -9/(8x^4)#

Explanation:

First of all, let's simplify the fraction to #3/(8x^3)#.
Two options:

Using the Quotient Rule

In case you didn't know, if #y = f(x)/g(x)#, then
#dy/dx = (f'(x)g(x) - g'(x)f(x))/(g(x)^2)#
In this case, #f(x) = 3# and #g(x) = 8x^3#. Therefore,
#dy/dx = ([3]' * (8x^3) - [8x^3]' * 3)/(8x^3)^2#.
#= (0 * (8x^3) - [8x^3]' * 3)/(8x^3)^2# (because #[3]' = 0#)
#= (-[8x^3]' * 3)/(8x^3)^2#
#= (-24x^2 * 3)/(8x^3)^2# (because #[8x^3]' = 24x^2#)
#= (-72x^2)/(64x^6)# (simplifying)
#= (-9x^2)/(8x^6)# (reducing)
#= -9/(8x^4)# (reducing exponents)

That is one way of doing it, but it's kind of hard. There is a simpler way of doing it.

Using the Power Rule

#y = 3/(8x)^3 = 3/8x^(-3)#
In case you didn't know, if #y = x^n#, then #dy/dx = n * x^(n-1)#. Therefore,
#dy/dx = 3/8 * (-3) * x^(-3-1)#
#= -9/8 * x^(-4)#
#= -9/(8x^4)#