- #1
Imperial
Hi all.
For any of you who have done differential calculus, I need a little help with a problem involving natural logarithms.
The question asks to differentiate y = ln x from first principles . It says "use the definition of the Euler number, namely e = lim(n->inf.) (1+1/n)^n.".
First principles means f'(x) = lim(h->0) [f(x+h) - f(x)] / h (this is the first thing we learned in calculus).
I so far managed two different methods:
Method 1. y = ln x
therefore e^y = x
dx/dy = e^y.
Since dx/dy * dy/dx = 1
1/(dx/dy) = dy/dx.
= 1/e^y
= 1/x.
Method 2. y = ln x
f(x) = ln x
f(x+h) = ln (x+h)
f'(x) = lim(h->0) [ln (x+h) - ln x] / h
= lim(h->0) ln (x+h/x) / h
= lim(h->0) 1/h * ln(1+h/x)
Since lim(h->0) ln(1+h/x) -> h/x where h != 0,
f'(x) = 1/h * h/x
= 1/x
Both of these methods work and are valid, although I didn't bring the definition of the Euler number into it. I personally have no idea how to do this. Could anyone here who has done a bit of math before please help me with this?
Thanks.
For any of you who have done differential calculus, I need a little help with a problem involving natural logarithms.
The question asks to differentiate y = ln x from first principles . It says "use the definition of the Euler number, namely e = lim(n->inf.) (1+1/n)^n.".
First principles means f'(x) = lim(h->0) [f(x+h) - f(x)] / h (this is the first thing we learned in calculus).
I so far managed two different methods:
Method 1. y = ln x
therefore e^y = x
dx/dy = e^y.
Since dx/dy * dy/dx = 1
1/(dx/dy) = dy/dx.
= 1/e^y
= 1/x.
Method 2. y = ln x
f(x) = ln x
f(x+h) = ln (x+h)
f'(x) = lim(h->0) [ln (x+h) - ln x] / h
= lim(h->0) ln (x+h/x) / h
= lim(h->0) 1/h * ln(1+h/x)
Since lim(h->0) ln(1+h/x) -> h/x where h != 0,
f'(x) = 1/h * h/x
= 1/x
Both of these methods work and are valid, although I didn't bring the definition of the Euler number into it. I personally have no idea how to do this. Could anyone here who has done a bit of math before please help me with this?
Thanks.