Question

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

Answer 1

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

Explanation:

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

• Use quotient rule
• Use power rule (easier)

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)`