We are asked to find the derivative of `g(x)=6x^2e^(-x) ` :

Use the product rule:

`g'(x)=(d/(dx)(6x^2))*e^(-x)+(6x^2)(d/(dx)(e^(-x)))`

`=(12x)e^(-x)+(6x^2)(-e^(-x)) ` (Using the derivative of `e^u ` )

` =e^(-x)(12x-6x^2) `

Find the derivative of `g(x)=6x^2*e^-x`

First off, you cannot derive immediately as there are two different variables multiplying each other. You will need to use the product rule which is `u'v+uv'`

This will give you: `g'(x)=( dy/dx*6x^2)(e^-x)+(6x^2)(dy/dx*e^-x)`

Which will then give you: `g'(x)=(12x)(e^-x)+(6x^2)(-e^-x)`

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`g(x) = 6x^2 e^-x`

The very first thing you should do is realize that this requires the product rule. The product rule is: `u'v + uv'` .

The "u" in this problem is `6x^2` , while the "v" is `e^-x` .

The derivative of "u" or u' is therefore `12x` because you bring down the 2 (and multiply it with the 6) and reduce the power by one.

The derivative of "v" or v' is `-e^-x` because you must use the chain rule. `-x` is also a function, so the derivative of `-x` is -1. The derivative of `e^-x` is itself multiplied by the inside function's derivative.

Now you can simply plug in all the variables. ``

`g'(x) = (12x)(e^-x) + (6x^2)(-e^-x)`

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## 6x^2e^-x

`6x^2 e^-x`

`12xe^-x - 6x^2 e^-x`

``This is product rule and chain rule all in one! Recall that the derivative of 6x^2 is just 12x. The derivative of e^-x uses the rule e^u => u'e^u so e^-x => -e^u. Put this together with the product rule and you have the answer.

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