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

2 Answers
Aug 16, 2017

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)" ")|)

Aug 16, 2017

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!