Question

find y'(x) of y = sin(e^2x) and find all solutions for y'(x) = 0.
I have that y'(x)= cos(e^2x)(e^2x)(2x) by chain rule. I don't get what finding solution's for
y'(x) = 0 means.

Answer 1

\[y'(x)=2 e^{2x}\cos(e^{2x})\]\[y'(x)=0 \implies \cos(e^{2x})=0 \] e^x is positive everywhere, so the only time the derivative will be 0 is when the cos(e^2x) part is equal to 0. So if that part is equal to 0, then the whole thing is equal to zero, so just find values of x in the cos(e^2x) = 0 expression. And that will give you where the derivative is 0.

@El_Tucan

Answer 2

still don't get it.

Answer 3

You get the derivative? @El_Tucan

Answer 4

yes

Answer 5

We have to find where the derivative is 0 right?

Answer 6

so plug in a zero

Answer 7

So we equate the derivative with 0 like:\[2e^{2x}\cos(e^{2x})=0\] Correct? Are you find with what I've done so far?

Answer 8

yes

Answer 9

So now to solve for x in (2e^2x)(cos(e^2x)) = 0, we must find out where it's 0. The derivative is 0, when one of these brackets is 0 correct? If one part 0, then the whole expression is 0, correct? Do you understand this so far?

@El_Tucan

Answer 10

because of multiplication?

Answer 11

Exactly. If one part is 0, then the whole product is 0. So for the whole thing to be 0, atleast one bracket needs to be 0 for the product to equal to 0 since we are multiplying. Do you understand that?

@El_Tucan

Answer 12

yes

Answer 13

Ok good.

Answer 14

So to find where (2e^2x)(cos(e^2x)) = 0, we need to find where one of the parts of the product equals 0. Because if one of the brackets is 0, then the product is 0. So we just need to solve for x in one of the brackets. But here is the deal:\[2e^{2x}>0, x \in \mathbb{R} \] The (2e^2x) part of the product is always a positive number for no matter what x I choose. Which means that this bracket will never equal 0. So for the whole expression to equal 0, we need atleast one bracket to equal to 0 to make the product 0. We know the first bracket with 2e^2x can never be zero and is always positive, so the only way the product can be 0 is when the second bracket, cos(e^2x) = 0. When this bracket equals 0, the whole expression equals 0. So by solving for x just in cos(e^2x) = 0, we can find for which x the whole product is 0. Do you understand now?

@El_Tucan

Answer 15

so the x has to make it zero

Answer 16

yes you have to find an x such that cos(e^2x) = 0, because when cos(e^2x) = 0, then the whole product of (2e^2x)(cos(e^2x) also equals 0.

Answer 17

sorry i didn't say thanks for the help.