How do you find the derivative of #y = (x  2)^3 (x^2 + 9)^4#?
1 Answer
Answer:
Explanation:
We're asked to find the derivative
#(dy)/(dx) [y = (x2)^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 = (x2)^3# 
#v = (x^2+9)^4# :
#y'(x) = (x^2+9)^4d/(dx) [(x2)^3] + (x2)^3d/(dx) [(x^2+9)^4]#
Now, we use the chain rule for the first term:
#d/(dx) [(x2)^3] = d/(du) [u^3] (du)/(dx)#
where

#u = x2# 
#d/(du) [u^3] = 3u^2# :
#y'(x) = (x^2+9)^4 3(x2)^2 d/(dx) [x2]+ (x2)^3d/(dx) [(x^2+9)^4]#
The derivative of
#y'(x) = 3(x^2+9)^4(x2)^2 + (x2)^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(x2)^2 + (x2)^2 4(x^2+9)^3d/(dx) [x^2+9]#
The derivative of
#color(blue)(ulbar(stackrel(" ")(" "y'(x) = 3(x^2+9)^4(x2)^2 + 8x(x2)^2 (x^2+9)^3" "))#