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!