Question

hey guys i need some clarification regarding the integration and differentiation of natural log and exponential functions...help?

Answer 1

like for example: how would you find the integral of \[x^{2x}\]

Answer 2

would it follow the integration rule of k^x where the integral is k^x/ln k times dx

Answer 3

general rules of integrals and derivatives of log and natural log would be very helpful

Answer 4

No. k^x formula is used ONLY when the constant is given, i.e. 3^x of 4^(2x).

Answer 5

yes k is a constant

Answer 6

But x^(2x) is NOT of the form (a^x), where a is a constant.

Answer 7

so which integral rule would be used to solve this?

Answer 8

when would i use the chain rule and when would i not?

Answer 9

Hmm I don't think a solution exists to such a problem..
Was it in your book or something..? :o

Example:\[\Large y\quad=\quad x^{2x}\]Taking the log of each side gives us,\[\Large \ln y\quad=\quad 2x\ln x\]Integrating each side with respect to it's proper variable gives us,\[\Large y \ln y- y\quad=\quad x^2 \ln x-\frac{1}{2}x+C\]And then we could ummmmm maybe... hmm

Answer 10

ummmmm is right
you are not going to find a nice closed form for this function

\[x^{2x}=e^{2x\ln(x)}\] and you can pretty much forget finding an anti derivative for it
you can find the derivative no problem, but not an anti derivative

Answer 11

yes its in my math book

Answer 12

i was actually just asking for general rules regarding the use of chain rules when integrating or find the derivative of the natural log and exponential functions...

Answer 13

Like these?\[\Large \int\limits \ln x, \qquad\qquad \int\limits a^x\;dx\]

Answer 14

because sometimes, as seen from my book, you would multiply the derivative of the inner function to ur outer functions and sometimes you wouldn't...

Answer 15

yea how those equal to 1/x and \[(a ^{x})/\ln a\]

Answer 16

respectively

Answer 17

and this is where my problem lies

Answer 18

for \[\int\limits_{?}^{?}a ^{x}dx\]

Answer 19

would it be \[(a ^{x})/\ln a\] and then multiplied by the derivative of the inner function x?

Answer 20

or just \[(a ^{x})/\]?

Answer 21

There isn't a chain rule for integration. At least I wouldn't describe it that way.

So you remember this derivative, yes?\[\Large (a^x)'\quad=\quad a^x \ln a\]

Answer 22

yes

Answer 23

and integral would be divide by ln a

Answer 24

So if we were to take this integral,\[\Large \int\limits \color{orangered}{a^x \ln a}\; dx\]What would we expect to get?

Answer 25

Yah good, it gives us back what we started with in that example.

Answer 26

I'm not sure what you mean by `multiply by the inner function x`.
Like if we had:\[\Large \int\limits a^{2x}\;dx\]Then yes, we get a little bit something more than just dividing by the ln(a).
Is that what you're asking about?

Answer 27

are you sure there isn't a chain rule for integration? i thought it applies to everything

Answer 28

yes so would u multiply it by 2 (the derivative of the inner function: 2x)? or not?

Answer 29

The `chain rule` (taking the derivative of a^2x) would produce an extra factor of 2.
So when we integrate, we need to divide by 2 to compensate for the extra factor of 2.

Answer 30

\[\Large (e^{2x})'\quad=\quad 2e^{2x}\]We ended up with a factor of 2 when differentiating,
So when we integrate, we need to divide by 2,\[\Large \int\limits e^{2x}dx\quad=\quad \frac{1}{2}e^{2x}+C\]

Answer 31

u forgot the 2 for the integral of e^2x that should've cancelled out the 1/2 on the right...

Answer 32

I was showing the difference between `differentiating e^2x` and `integrating e^2x` but yes you have the right idea :)

Answer 33

so the chain rule apply to which? sry i suck at this :P

Answer 34

This works quite nicely when our `inner function` is linear.
If it's linear with a coefficient, we simply divide by that coefficient when we integrate.

Unfortunately, unlike with taking a derivative, this trick doesn't always work.
If the inner function is NOT linear, we have something else going on.

Example:\[\Large \left(e^{x^2}\right)'\quad=\quad (2x)e^{x^2}\]So taking a derivative produces a 2x. Integrating the same function:\[\Large \int\limits e^{x^2}\;dx\]does not allow us to simply divide by the derivative of the inner function (2x).

Answer 35

Mmm maybe I'm making it more complicated than it needs to be.. hmm thinking..

Answer 36

lol

Answer 37

actually

Answer 38

for integration u would divide it by the derivative of the inner function