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