Question

How do you differentiate `f(x)= (3x+1)^4(2x-3)^3` using the product rule?

Answer 1

`12(3x+1)^3(2x-3)^3 + 6(2x-3)^2(3x+1)^4`

Explanation:

This problem actually involves two rules: the product rule, and the chain rule. But first, let's just focus on the product rule. The general equation for this is:

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

Now, we just plug in everything:

`d/dx[(3x+1)^4]` `(2x-3)^3` `+ d/dx[(2x-3)^3]` `(3x-1)^4`

The problem arises at the fact that your two functions are actually compositions of functions. For example, `(3x+1)^4` is really the composition of the two functions `x^4` and `3x+1`. Hence, we'll need to use the chain rule to evaluate these derivatives. Let's take each one in turn:

`d/dx[(3x+1)^4]`

So first, we do the outermost function (in this case `x^4`)'s derivative, then multiply by the derivative of the inner function (`3x+1`). So, this gives us:

=> `4(3x+1)^3 * d/dx(3x+1)`
=> `4(3x+1)^3 * 3`
=> `12(3x+1)^3`

Now, for the second one:

`d/dx[(2x-3)^3]`

Again, same process. Derivative of the outside, then multiply by derivative of the inside:

=> `3(2x-3)^2 * d/dx(2x-3)`
=> `3(2x-3)^2 * 2`
=> `6(2x-3)^2`

We're done! Now, we just go ahead and plug these back into the product rule equation we got for our final answer:

`12(3x+1)^3(2x-3)^3 + 6(2x-3)^2(3x+1)^4`

You could foil out some stuff here, but in my opinion, it's really just more work. Easier to leave it as is.

Hope that helped :)