# Calculus /Derivatives of e^x and ln e

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**Question**

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

is 1)

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.comCategory: Bank

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