Question

How do you find the derivative of `(1+x+[3x^2])^4`?

Answer 1

`4*(1+6x)(1+x+3x^2)^3`

Explanation:

`((f(x))^n)dx/dy=n*f'(x)*f(x)^(n-1)`

`((1+x+[3x^2])^4)dx/dy=4*(1+6x)(1+x+3x^2)^3`

Answer 2

`dy/dx ->f'(x)=4(1+6x)(1+x+3x^2)^3`

Explanation:

Using the 'old fashioned' notation

Given: `y=(1+x+3x^2)^4`

Set `u=1+x+3x^2color(white)("d")` then `color(white)("d")(du)/dx=1+6x`

Also we have `y=u^4 color(white)("d")->color(white)("d")dy/(du)=4u^3`
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Really the format `dy/dx` is a fraction and it behaves the same way fractions do. Consequently we can multiply them.

Observe that: `(du)/dx xx dy/(du)color(white)("d") =color(white)("d") (du)/(du)xxdy/dx color(white)("d")=color(white)("d") dy/dx`

Using the above:

`dy/dx = (1+6x)xx4u^3`

`dy/dx = (1+6x)xx4(1+x+3x^2)^3`

`dy/dx ->f'(x)=4(1+6x)(1+x+3x^2)^3`