Question

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

Answer 1

`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))`