Why is the indefinite integral of e^6x not the same as e^6x?

[DeepSeeded]

I was under the impression that the derivative of e raised to any power with a variable in it is just the same result. So why isn't it the same for all indefinate integrals of e to any power with a variable in it? Specificaly why is it that:

\int e^6x = \frac{1}{6} e^6x

 

[Vid]

I'm assuming you mean e^(6x).

It's a simple u-sub. u = 6x. du = 6dx du/6 = dx

Then Integral[e^(6x)dx] becomes Integral[e^(u)*dx/6] = 1/6*e^(u) = 1/6*e^(6x)

 

[DeepSeeded]

I'm assuming you mean e^(6x).

It's a simple u-sub. u = 6x. du = 6dx du/6 = dx

Then Integral[e^(6x)dx] becomes Integral[e^(u)*dx/6] = 1/6*e^(u) = 1/6*e^(6x)

Ok that makes sense. However the integral is an anti-derivative. This means that now taking the derivative of the answer:

F = 1/6*e^(6x)
F' = e^(6x)

Is that true?

I thought it was F' = 1/6*e^(6x)

 

[sutupidmath]

Ok that makes sense. However the integral is an anti-derivative. This means that now taking the derivative of the answer:

F = 1/6*e^(6x)
F' = e^(6x)

Is that true?

I thought it was F' = 1/6*e^(6x)

yeah, when you take the derivative of the primitive function F(x) it should equal the integrand. that is

\int f(x)=F(x)+C=>[F(x)+C]'=f(x)

Integrals and derivatives, in some sense are inverse to each other, that is one undoes what the other does.

 

[DeepSeeded]

yeah, when you take the derivative of the primitive function F(x) it should equal the integrand. that is

\int f(x)=F(x)+C=>[F(x)+C]'=f(x)

Integrals and derivatives, in some sense are inverse to each other, that is one undoes what the other does.

Ok I was wrong about e, e raised to a single variable with no coefficient only comes out to be the same answer.
Otherwise it comes out to be the product of the coefficient and the same answer. Thats what threw me off to begin with. Thanks

 

[Mute]

Ok I was wrong about e, e raised to a single variable with no coefficient only comes out to be the same answer.
Otherwise it comes out to be the product of the coefficient and the same answer. Thats what threw me off to begin with. Thanks

In general you have to use the chain rule:

\frac{d}{dx}e^{f(x)} = e^{f(x)}\frac{d f(x)}{dx}

When f(x) = x you get the result you were originally familiar with.

 

[HallsofIvy]

I was under the impression that the derivative of e raised to any power with a variable in it is just the same result.

This is incorrect. The derivative of e^x is e^x. If the exponent just "has a variable in it", i.e. is a function of x, they you must use the chain rule:
\frac{de^{f(x)}}{dx}= e^{f(x)}\frac{df}{dx}

 

[tectactoe]

I used to have much confusion about this as well, but always remember your integral/derivative rules and check your answers.

In this case, \int e^{6x}dx = \frac{1}{6}e^{6x} + C

To check, take the derivative of \frac{1}{6}e^{6x} + C and you should come back to e^{6x}.

The \frac{1}{6} is pulled through the derivative sign by constant multiple rule and the C drops out, so you're dealing with

\frac{d}{dx}(e^{6x}) which follows the rule \frac{d}{dx}(e^{u}) = e^{u}\frac{du}{dx}

and \frac{d}{dx}(6x) = 6

So, \frac{d}{dx}(e^{6x}) gives you 6e^{6x}

and don't forget about the constant 1/6 you pulled out earlier...

(\frac{1}{6})(6e^{6x}) = e^{6x}

It checks out.