How Do You Differentiate e^x and ln(x)?

In summary, to differentiate y=e^x, we use the rule for differentiating the inverse of a function since the exponential function is the inverse of the natural logarithmic function. This gives us the derivative of e^x as e^x. For y=lnx, we can use implicit differentiation with the help of exponential function to find the derivative, which is equal to 1/x. The process for proving these derivatives depends on how the functions are defined. One way is to use the difference quotient and extend the definition to all real numbers, while another way is to use the Fundamental Theorem of Calculus and define ln(x) as the inverse of exp(x).
  • #1
JasonX
10
0
please explain in more detail on how we come up with the answers below. Thanks in advance!
(formulas much appreciated)

Differentiate:
1.
y=e^x
=e^x
2.
y=lnx
=1/x
 
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  • #2
There are many websites on the internet explaining those two.

Just search for the derivative for the first one though. Because for the second one, you can use implicit differentiation as follows...

y = lnx (use exponential)

e^y = x (differentiate implicitly)

y' e^y = 1

y' = 1/(e^y) (now sub in y)

y' = 1/(e^(lnx)) = 1/x
 
  • #3
The first one is found using the rule for differentiating the inverse of a function,

[tex](f^{-1})^{'}(x) = \frac{1}{f^{'}(f^{-1})} [/tex]

because the exponential function is the inverse of the natural logarithmic function.

so

[tex]\frac{d}{dx}e^{x} = \frac{1}{\frac{d}{dx}[ln (e^{x})]} [/tex]

[tex]\frac{d}{dx}e^{x} = \frac{1}{\frac{1}{e^{x}}}[/tex]


[tex]\frac{d}{dx}e^{x} = e^{x}[/tex]
 
  • #4
One point I think should be made: watch what you say! In particular, watch where you put "=". The most natural interpretation of
y=lnx
=1/x
is that you are saying that y is equal to both ln x and 1/x- so they are equal to each other.

Be very careful that you say
y= ln x,
y'= ln x.

As to how you prove them, that depends strongly on how you define the functions!
If you define ax by first defining an= "product of a with itself n times" as long as n is a positive integer, then extend to all rational numbers by requiring that ax+y= axay and axy= (ax)y for x,y any rational numbers, and finally extending to any real number by "continuity", then you can use the "difference quotient":
[tex]\lim_{h\rightarrow 0}\frac{a^(x+h)-a^x}{h}= \lim_{h\rightarrow 0}\frac{a^xa^h- a^x}{h}= \lim_{h\rightarrow 0}\frac{a^h- 1}{h} a^x[/itex]
You can prove that [itex]\lim_{h\rightarrow 0}(a^h- 1)/h[/itex] exists for a any positive number, so that the derivative of [itex]a^x[/itex] is a constant times [itex]a^x[/itex] and then define e to be the number such that that constant is 1.
Having done that, define ln(x) to be the inverse function to [itex]e^x[/itex]. If y= ln(x) then x= [itex]e^y[/itex] so [itex]dx/dy= e^y= x[/itex] and then [itex]dy/dx= 1/x[/itex].

Conversely, it has become common in calculus books to define ln(x) as [itex]ln(x)= \int_1^x (1/t)dt[/itex] so that we have immediately from the Fundamental Theorem of Calculus that d(ln x)/dx= 1/x. Then define exp(x) to be the inverse function to ln(x) and we have: if y= [itex]e^x[/itex] then x= ln(y) so dx/dy= 1/y and dy/dx= y= [itex]e^x[/sup].
 
  • #5
HallsofIvy said:
y= ln x,
y'= ln x

I'm afraid that the OP may lose the point you were trying to make if you leave this as it is, otherwise I wouldn't bother correcting something so obvious.

Also you put parentheses where you meant to put braces.
 

FAQ: How Do You Differentiate e^x and ln(x)?

What is the definition of e^x?

The number e is a mathematical constant that is approximately equal to 2.71828. The function e^x is the exponential function with base e, where x is the exponent.

What is the definition of ln x?

The natural logarithm function, ln x, is the inverse of the exponential function e^x. It is defined as the power to which e must be raised to equal x.

How do you differentiate e^x?

The derivative of e^x is equal to e^x. In other words, the rate of change of e^x is always equal to the value of e^x.

How do you differentiate ln x?

The derivative of ln x is equal to 1/x. In other words, the rate of change of ln x is inversely proportional to x.

What is the relationship between e^x and ln x?

e^x and ln x are inverse functions. This means that e^x and ln x undo each other, and the compositions of the two functions result in the original input. In other words, e^ln x = x and ln(e^x) = x for all real numbers x.

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