Question

How do you use the chain rule to differentiate `y=1/(x^4+x)^2`?

Answer 1

`color(blue)(y'(x) = (-2(4x^3+1))/((x^4+x)^3)`

Explanation:

We're asked to find the derivative

`(dy)/(dx) [y = 1/((x^4 + x)^2)]`

We can first use the quotient rule, which is

`d/(dx) [u/v] = (v(du)/(dx) - u(dv)/(dx))/(v^2)`

where

• `u = 1`

• `v = (x^4 + x)^2`:

`y'(x) = ((x^4+x)^2d/(dx)[1] - 1d/(dx) [(x^4+x)^2])/((x^4+x)^4)`

The derivative of `1` (a constant) is `0`, leaving us with

`y'(x) = (-1d/(dx) [(x^4+x)^2])/((x^4+x)^4)`

Now, we'll use the chain rule:

`d/(dx) [(x^4+x)^2] = d/(du) [u^2] (du)/(dx)`

where

• `u = x^4 + x`

• `d/(du) [u^2] = 2u` (from power rule):

`y'(x) = (-2(x^4+x)d/(dx)[x^4+x])/((x^4+x)^4)`

The derivative of `x^4 + x` is `4x^3 + 1` (from power rule):

`y'(x) = (-2(x^4+x)(4x^3 + 1))/((x^4+x)^4)`

Which simplifies to

`color(blue)(ulbar(|stackrel(" ")(" "y'(x) = (-2(4x^3+1))/((x^4+x)^3)" ")|)`

Answer 2

`y' = -(8x^3 + 2)/(x^4 + x)^3`

Explanation:

Alternatively, we could rewrite as

`y = (x^4 + x)^-2`

Letting `u = x^4 + x` and `y = u^-2`, we can use the chain rule to obtain:

`y' = -2u^-3 * 4x^3 + 1`

`y' = -2(x^4 + x)^(-3) * 4x^3 + 1`

`y' = -(8x^3 + 2)/(x^4 + x)^3`

Hopefully this helps!