Chain Rule/Product Rule Question

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In summary, the problem involves finding the value of x when f'(x)=0 for the given function f(x) = (6-x2) * e2x. The conversation discusses using both the Product Rule and the Chain Rule to find the derivative of e2x, with the correct approach being to use the Chain Rule. The final solution involves solving for x using the derived equation.
  • #1
Mattara
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I am suppose to derive the following and then find x if f´(x)= 0

f(x) = (6-x2) [tex]\bullet[/tex] e2x

Which one should be used? I guess I am confused with the fact that it is e2x

Product rule gives:

(6-x2) [tex]\bullet[/tex] e2x + e2x [tex]\bullet[/tex] (-2x)

e2x(6-x2-2x) (x1 = 0)

-x2-2x+6 = 0 gives decimal values that doesn't seem to fit.

Chain rule gives:

I'm not even sure how to start here. I assume that the outer derivate is e2x? Or should the chain rule be applied to it and then the product rule?

Some kind of hint as to what is the correct path would be greatly appreciated :smile:

Edit: Latex mishap >_>
 
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  • #2
I assume that the big, big dot in your post is the multiple sign.
You can use both rules (i.e, Chain Rule, and Product Rule) in this problem. There's no limit of the number of the rules you can use. So, just use it where you think is appropriated.
f(x) = (6 - x2) e2x
f'(x) = (6 - x2)' e2x + (6 - x2) (e2x)' = ...
You have taken the derivative of e2x with respect to x incorrectly.
We should use the Chain Rule there. By letting u = 2x, we have:
[tex]\frac{d}{dx} e ^ {2x} = \frac{d (e ^ u)}{du} \times \frac{du}{dx} = \frac{d(e ^ u)}{du} \times \frac{d(2x)}{dx} = ...[/tex]
So, can you see where your mistake is?
Can you go from here? :)
 
  • #3
Of course.

The inner is 2.

2e2x(6-x2)+ e2x(-2x)

e2x(12-2x2)+ e2x(-2x)

12-2x2-2x = 0

and there we have the values of x.

Thank you VietDao29, you were very helpful :smile:
 

1. What is the Chain Rule?

The Chain Rule is a formula used in calculus to find the derivative of composite functions. It states that the derivative of a composite function is equal to the derivative of the outer function multiplied by the derivative of the inner function.

2. How do you use the Chain Rule to find the derivative of a function?

To use the Chain Rule, you first identify the outer function and the inner function. Then, you take the derivative of the outer function and multiply it by the derivative of the inner function. It can also be written as d/dx[f(g(x))] = f'(g(x)) * g'(x).

3. What is the Product Rule?

The Product Rule is another formula used in calculus to find the derivative of two functions that are multiplied together. It states that the derivative of a product of two functions is equal to the first function times the derivative of the second function plus the second function times the derivative of the first function.

4. How do you use the Product Rule to find the derivative of a function?

To use the Product Rule, you first identify the two functions that are being multiplied together. Then, you take the derivative of the first function and multiply it by the second function, and add it to the derivative of the second function multiplied by the first function. It can also be written as d/dx[f(x)g(x)] = f'(x)g(x) + g'(x)f(x).

5. What is the difference between the Chain Rule and the Product Rule?

The Chain Rule is used to find the derivative of composite functions, while the Product Rule is used to find the derivative of two functions that are multiplied together. The Chain Rule involves taking the derivative of the inner function and then multiplying it by the derivative of the outer function, while the Product Rule involves multiplying the two functions together and then taking the derivative. They are two separate rules that are used in different situations.

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