Calculus /Derivatives of e^x and ln e
My question regards the different results obtained in textbooks when one finds the derivative of e to the x and one finds the derivative of ln e (or e to the y).
Discussion of problem:
1. derivative of e^x = e^x times lin e = e^x times 1 (since lin e = 1)
or the proof goes like this:
2. derivative of e^x = e^x times x' (that is, e to the x times the derivative of x) = e^x times 1
Thus, the textbooks always show that the derivative of e^x is e^x.
However, when it comes to finding the derivative of lin x, the procedure seems to be different.
First, it is pointed out that lin x is the same as e^y = x. Typically, the textbooks then go on
to find the derivative of ey as follows:
e^y = x
so the derivative of e^y = e^y times lin e times y' (HERE IS THE PROBLEM, the text books stick in this extra y', sometimes you see this in the form dy/dx)
x' = e^y times lin e times y'
and (since the derivative of x
Well. your textbook seems to be confusing you on a simple proof. Here is the argument,
Let y = lnx.
Then e^y = x.
Take the derivative of both sides with respect to the variable x. That means we treat y as a function of x. This gives
y' e^y = 1
Because y = lnx. e^y = e^lnx = x
This means we can substitute x for e^y in y' e^y = 1 .
This gives y' x = 1
Then y' = 1/x. which is the result you want.
You must understand that the derivative of e^y is not e^y when the derivative is taken with respect to x. The derivative of e^y is e^y when the derivative is taken with respect to y.
Have you seen the rule (e^f(x))' = f'(x) e^f(x).
That is the rule we are using when we get y'e^y as the derivative of e^y. You must see that y is a function of x and is treated as a function. not as a single. independent variable.
I hope this helps. I would need to see your book to try to explain further.Source: en.allexperts.com