Question

integral of (x^2)*(3^x^3)(dx). I get to use u substitution with u=x^3, but I get stuck when trying to integrate 3^u.

Answer 1

what's the integral of e^x?

what's the derivative of e^x?

that should get you thinking

Answer 2

They're both e^x, but this isn't e, it's a constant.

Answer 3

is 3 not a constant?

Answer 4

Do you mean is e not a constant?

Answer 5

think of why the derivative of e^x is still e^x

Answer 6

I thought that was some special property of e.

Answer 7

not exactly

the derivative of a^x is log(a)*a^x where a is a constant

log(e) = 1 that's why the derivative of e^x is just e^x

Answer 8

Okay, I found that the integral of a^xdx is (a^x)/log(a), and I kind of see how that's related to the derivative function where the derivative is multiplication and the integral's division, but I can't quite grasp why they are what they are.

Answer 9

You can do a `change of base` on the exponential to convince yourself.
That is certainly a good way to see what's going on.

I would recommend however that you just try to get used to integration.

For example:
\(\large (e^{2x})'=2e^{2x}\)
Taking the derivative produces a factor of 2 right?
\(\large \int 2e^{2x}dx\)
What happens when we integrate this?
We end up LOSING a factor of 2, correct?
Another way to think of this is that we're actually DIVIDING by 2 instead of multiplying.\[\large \int\limits\limits 2e^{2x}dx=2\frac{1}{2}e^{2x}=e^{2x}\]

Try to get used to this idea. If a derivative would produce a constant coefficient, then the integral of the same term will produce the reciprocal of that coefficient.

\(\large (3^x)'=3^x (\ln 3)\)

Therefore following the same logic as before,
\(\large \int 3^x=\dfrac{1}{\ln 3}3^x\)

Answer 10

Oh maybe that's not the part you were confused on :\ my bad.

Are you asking, Why does the derivative of \(\large 3^x\) produce \(\large 3^x(\ln x)\)?

Answer 11

ln 3* woops, typo

Answer 12

I was confused about both things, actually.

Answer 13

The first part makes sense though, thanks for that.

Answer 14

Rob explained this a little bit already, but the idea with exponential functions is that the derivative always produces a factor of natural log of the base.

\[\large y=3^x\]Taking the natural log of both sides,\[\large \ln y=\ln 3^x\]Applying a rule of logarithms,\[\large \ln y = x \ln 3\]
Taking the derivative of both sides with respect to x gives us,\[\large \frac{y'}{y}=\ln 3\]Lemme know if that step is confusing, there's a bit of chain rule going on there.
\[\large y'=\color{green}{y}(\ln 3)\]From here, let's recall what our \(\large y\) was, \(\large \color{green}{y=3^x}\)
Which gives us a derivative of,\[\large y'=\color{green}{3^x}(\ln 3)\]

Answer 15

This also ends up happening with e^x.
But since ln e =1 we usually don't mention it.

\[\large y=e^x\]\[\large y'=e^x(\ln e) \qquad \rightarrow \qquad y'=e^x(1)\]

Answer 16

Sorry for the delay, I'm at work and we got busy. Yeah, the chain rule when you take the derivative is definitely confusing me. The derivative of lny is 1/y, so the left side makes sense, but the right side, not so much.

Answer 17

Right side not so much? Yah people tend to get confused by these things sometimes.

It's a trick! Don't be fooled.

See that ln3? It's just a number.
It's a fancy looking constant, dressed up in a tuxedo. He's just trying to fool you.

What is the derivative of \(\large 3x\)?
It gives us \(\large 3(1)\) or simply, \(\large 3\).

So what's the derivative of \(\large (\ln 3)x\)?
Remember, it's just a constant coefficient!
Following the same rule we used before, gives us, \(\large (\ln 3)(1)\) or simply \(\large \ln3\)

This type of thing comes up every once in a while, so make sure you're comfortable with it. Your teacher might throw you a curve ball and ask you what the derivative of \(\large e^{\ln3}\) is.
Your first instinct is to treat it like e^x, but notice that there is no variable anywhere in the problem. It's just a constant! :O