Derivatives of natural numbers

[apples]

1.
I was trying to understand the proof of
(d/dx) b^x = (ln b)*(b^x)
it says:
b= e^(ln b)
so, b^x= e^((ln b)*x))
So now we use the chain rule:
(d/dx) b^x = (d/dx) e^((ln b)*x))

I understand everything so far, but not the next step.
It says then that
(d/dx) e^((ln b)*x))= (ln b)*e^(ln b)*x)

how did they get this, i am confused about the bold part. I am forgetting something about taking the derivative of the natural number 'e'?
I know that d/dx e^x = x

so how does the ln b come behind e^x above?

what does (d/dx) e^f(x) equal to?

is (d/dx) e^f(x) = f '(x)* e^f(x) ?
if so then
1 + ln b
should come behind that because
d/dx (ln b)*x is equal to 1 + ln b


Please help me
thank you

 

[apples]

uhh

hello?

 

[apples]

ok i found out that
(d/dx) e^f(x) = f '(x)* e^f(x)

still i am confused about the other thing.

 

[Dick]

d/dx(ln(b)*x)=ln(b). Not ln(b)+1. Why would you think it's that? b is a constant. d/dx(ln(b))=0.

 

[apples]

Because when you use the product rule on
d/dx (ln(b) * x) you get (1 + ln (b))

 

[Dick]

d/dx(ln(x)*x)=(1+ln(x)). There's a b in there. Not two x's.

 

[apples]

so could you please tell me how to solve
d/dx (ln(b) * x)

 

[Dick]

d/dx(ln(b))=(d/dx(b))/b. If b is a constant then d/dx(b)=0, right? d/dx(ln(b)*x)=ln(b).

 

[apples]

Wow, I'm so stupid.

Thanks a lot! You saved my day!

 

[Integral]

so could you please tell me how to solve
d/dx (ln(b) * x)

ln(b) is a constant.

[tex] \frac {d} {dx} ln(b) x = ln(b) \frac {dx} {dx} = ln(b) [/tex]

 

[boombaby]

look at the definition of derivatives and you will surely understand how does that fomula work