- #1

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^{2x}, x ε ℝ And determine the domain.

So I did...

f'(x) = xe

^{2x}[itex]\bullet[/itex] d/dx 2

I applied the chain rule. I'm not sure if I did this right.

- Thread starter CrossFit415
- Start date

- #1

- 160

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So I did...

f'(x) = xe

I applied the chain rule. I'm not sure if I did this right.

- #2

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you'll have to use the product rule AND the chain rule

- #3

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Ahh I see thank you

- #4

Mark44

Mentor

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And in that order.you'll have to use the product rule AND the chain rule

- #5

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Okay cool!

- #6

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So I applied the product rule 1st;

f(x) = xe^{2x}

f'(x) = xe(2x) + xe(2)

Did I do this correctly?

f(x) = xe

f'(x) = xe(2x) + xe(2)

Did I do this correctly?

- #7

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First, what is the derivative of e^(2x) ?

Second, if f and g are arbitrary functions of x, what is the derivative of f*g with respect to x (ie, what does the product rule say)?

- #8

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derivative of e^{2x} is 2e?

- #9

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Nope !

Here we have to use the chain rule but before we get there, what is the derivative of e^(x)?

Here we have to use the chain rule but before we get there, what is the derivative of e^(x)?

- #10

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Just e^{x}

So...

e^{2x} = 2e^{x} ?

So...

e

- #11

- 164

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Maybe if you saw another example it would become more clear...

What is the derivative of sin(2x)?

Or if you prefer by the definition of the chain rule:

[itex](f \circ g)' = f'(g) * g'[/itex]

In your case, what is f and what is g?

Then can you see your mistake?

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