# Thread: e^x containing composite function

1. ## e^x containing composite function

I'm wondering...With $\displaystyle e^x$

Why doesn't $\displaystyle \ln x$ count as an "inside" function?

Example:

$\displaystyle \frac{dy}{dx}e^{sin(x)}=e^{sin(x)}cos(x)$

Here we see an example of the chain rule. Derivative of the inside function times the derivative of the outside function...

However, with this example:

$\displaystyle \frac{dy}{dx}e^{\ln 7x}=e^{\ln 7x}\cdot \ln 7$

There is no derivative of the inside and then outside. It seems to adhere to normal $\displaystyle e^x$ differentiation rules such as:

$\displaystyle \frac{dy}{dx}e^{7x}=7\cdot e^{7x}$

Why is this? Thanks!

2. ## Re: e^x containing composite function

Originally Posted by Paze
example:

$\displaystyle \frac{dy}{dx}e^{\ln 7x}=e^{\ln 7x}\cdot \ln 7$

There is no derivative of the inside and then outside. It seems to adhere to normal $\displaystyle e^x$ differentiation rules such as:

$\displaystyle \frac{dy}{dx}e^{7x}=7\cdot e^{7x}$
I simply do not understand what you are on about.

First $\displaystyle e^{\ln 7x}=7x$

so $\displaystyle \frac{dy}{dx}e^{\ln 7x}=7$.

3. ## Re: e^x containing composite function

Originally Posted by Plato
I simply do not understand what you are on about.

First $\displaystyle e^{\ln 7x}=7x$

so $\displaystyle \frac{dy}{dx}e^{\ln 7x}=7$.
Sorry. It's supposed to say $\displaystyle e^{(\ln 7)\cdot x}$

Do you understand now?

4. ## Re: e^x containing composite function

That's because the derivative of u = sinx is more complicated than the derivative of u = cx for any constant c. in your example the constant is c = ln (7).

5. ## Re: e^x containing composite function

Originally Posted by agentmulder
That's because the derivative of u = sinx is more complicated than the derivative of u = cx for any constant c. in your example the constant is c = ln (7).

So logarithms aren't functions?

6. ## Re: e^x containing composite function

Originally Posted by Paze
So logarithms aren't functions?
Of course not; functions map a range of inputs to a range of outputs, so therefore require at least a single variable to do so. So for instance e^x is a function that maps x to e^x. e^5 is not a function, it always has constant value. Ln(7x) is a function, but ln(7) is not. The derivative (wrt to x) of e^(ln7x) is as has been pointed out, 7, but the derivate of e^((ln7)x) is ln(7)e^((ln7)x) since it is of the form e^kx where k=ln7 so you can see it does indeed follow normal differentiation laws.

7. ## Re: e^x containing composite function

Originally Posted by Nemesis10192
Of course not; functions map a range of inputs to a range of outputs, so therefore require at least a single variable to do so. So for instance e^x is a function that maps x to e^x. e^5 is not a function, it always has constant value. Ln(7x) is a function, but ln(7) is not. The derivative (wrt to x) of e^(ln7x) is as has been pointed out, 7, but the derivate of e^((ln7)x) is ln(7)e^((ln7)x) since it is of the form e^kx where k=ln7 so you can see it does indeed follow normal differentiation laws.
Right. Thanks!

8. ## Re: e^x containing composite function

Uuhhh... it depends on what we are talking about... is a single 1 dimensional number a function ? I don't know. In 2 dimensions f(x) = c is a function , it is a horizontal line that takes on all x values exactly once and every x value has only 1 y value and therefore passes the so called 'vertical line test'. Similarly f(x) = ln 7 is a function in 2 (or more) dimensions. For any real number c , one may decide to rewrite this as

$\displaystyle f(x) = cx^0$

In order to have a 'place' to sbstitute for x ... this works out fine for all real x EXCEPT x = 0 where we encounter the 'disagreeable'

$\displaystyle f(0) = c \cdot 0^0$

Personally, it doesn't bother me too much and i won't throw it away just because there is a problem at x = 0 , but that's me , others may disagree.