What is the derivative of #(1+4x)^5(3+x-x^2)^8#?

Question

Christopher Agresta · Accepted Answer

#y^' = 4(1 + 4x)^4 * (3 + x - x^2)^7 * (-21x^2 + 9x + 17)#

You can differentiate this function by using the chain rule and the product rule.

Notice that your function can be written as

#y = f(x) * g(x)#

which means that its derivative can be determined using the product rule

#d/dx(y) = [d/dx(f(x))] * g(x) + f(x) * d/dx(g(x))#

In your case, #f(x) = (1+4x)^5# and #g(x) = (3 + x - x^2)^8#.

Differentiate these functions separately by using the chain rule. For #f(x)#, you have

This will get you

#d/dx(u^5) = d/(du)(u^5) * d/dx(u)#

#d/dx(u^5) = 5u^4 * d/dx(1+4x)#

#d/dx(1+4x)^5 = 5 * (1 + 4x)^4 * 4#

For #g(x) you have

This will get you

#d/dx(u_1^8) = d/(du_1)(u_1^8) * d/dx(u_1)#

#d/dx(u_1^8) = 8u_1^7 * d/dx(3 + x - x^2)#

#d/dx(3 + x - x^2)^8 = 8(3 + x - x^2)^7 * (1 - 2x)#

Take these derivatives to your target calculation to get

#y^' = 20 * (1 + 4x)^4 * (3 + x - x^2)^8 + (1 + 4x)^5 * 8 * (1-2x) * (3 + x - x^2)^7#

#y^' = 4(1 + 4x)^4 * (3 + x - x^2)^7 * [ 5 * (3 + x - x^2) + 2 (1-2x)(1 + 4x)]#

This is equivalent to

#y^' = 4(1 + 4x)^4 * (3 + x - x^2)^7 * [15 + 5x + 5x^2 + 2(1 + 2x - 8x^2)]#

#y^' = 4(1 + 4x)^4 * (3 + x - x^2)^7 * (15 + 5x - 5x^2 + 2 + 4x - 16x^2)#

#y^' = color(green)(4(1 + 4x)^4 * (3 + x - x^2)^7 * (-21x^2 + 9x + 17))#