Question

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

Answer 1

You have two options: The first is to expand that polynomial and take the derivative. The other is to do a substitution to get the solution: `(du)/(dx)=4(x^2+3x+1)^3*(2x+3)`

Explanation:

I'm going to go with the second option, because I'm not a masochist (jokes!)

When you take a derivative of a set of terms in parentheses, you can do a substitution, treating the parentheses as a separate single function. We'll treat the in-parentheses terms like this:

`(x^2+3x+1)=g(x)`

so our function now looks like this:

`u=g(x)^4`

Now, the derivative is not simply `(du)/(dx)=4g(x)^3`, but it is instead `(du)/(dx)=4g(x)^3*(dg)/(dx)`. We must calculate `(dg)/(dx)` before arriving at our solution.

if `g(x)=x^2+3x+1` then:

`(dg)/(dx)=2x+3`

Re-inserting our substitutions, we now arrive at our answer:

`(du)/(dx)=4(x^2+3x+1)^3*(2x+3)`

Answer 2

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

Explanation:

Use the chain and power rule. Find the derivative of `1(....)^4`(use the power rule. Then find the derivative of `(x^2+3x+1)`(use the power rule). then combine the two. `4(x^2+3x+1)^3(2x+3)`

Answer 3

`(du)/dx=4(2x+3)(x^2+3x+1)^3`

Explanation:

`"differentiate using the "color(blue)"chain rule"`

`"Given "y=f(g(x))" then"`

`dy/dx=f'(g(x))xxg'(x)larrcolor(blue)"chain rule"`

`u=(x^2+3x+1)^4`

`rArr(du)/dx=4(x^2+3x+1)^3xxd/dx(x^2+3x+1)`

`color(white)(rArr(du)/dx)=4(2x+3)(x^2+3x+1)^3`