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

1 Answer
Aug 11, 2015

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

Explanation:

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

  • f(u) = u^5, with u = (1 + 4x)

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

  • g(u_1) = u_1^8, with u_1 = 3 + x - x^2

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))