How do you find the derivative of #y = (x - 2)^3 (x^2 + 9)^4#?

1 Answer
Aug 6, 2017

Answer:

#y'(x) = 3(x^2+9)^4(x-2)^2 + 8x(x-2)^3 (x^2+9)^3#

Explanation:

We're asked to find the derivative

#(dy)/(dx) [y = (x-2)^3(x^2+9)^4]#

Let's first use the product rule, which is

#d/(dx) [uv] = v(du)/(dx) + u(dv)/(dx)#

where

  • #u = (x-2)^3#

  • #v = (x^2+9)^4#:

#y'(x) = (x^2+9)^4d/(dx) [(x-2)^3] + (x-2)^3d/(dx) [(x^2+9)^4]#

Now, we use the chain rule for the first term:

#d/(dx) [(x-2)^3] = d/(du) [u^3] (du)/(dx)#

where

  • #u = x-2#

  • #d/(du) [u^3] = 3u^2#:

#y'(x) = (x^2+9)^4 3(x-2)^2 d/(dx) [x-2]+ (x-2)^3d/(dx) [(x^2+9)^4]#

The derivative of #x-2# is #1#:

#y'(x) = 3(x^2+9)^4(x-2)^2 + (x-2)^3d/(dx) [(x^2+9)^4]#

Now we use the chain rule on the second term:

#d/(dx) [(x^2+9)^4] = d/(du) [u^4] (du)/(dx)#

where

  • #u = x^2+9#

  • #d/(du) [u^4] = 4u^3#:

#y'(x) = 3(x^2+9)^4(x-2)^2 + (x-2)^2 4(x^2+9)^3d/(dx) [x^2+9]#

The derivative of #x^2+9# is #2x#:

#color(blue)(ulbar(|stackrel(" ")(" "y'(x) = 3(x^2+9)^4(x-2)^2 + 8x(x-2)^2 (x^2+9)^3" ")|)#