# How can I find the derivative of #y=e^x# from first principles?

##### 1 Answer

# d/dx e^x = e^x #

#### Explanation:

We seek:

# d/dx e^x#

**Method 1 - Using the limit definition:**

# f'(x) = lim_{h to 0} {f(x+h)-f(x)}/{h} #

We have:

# f'(x) = lim_{h to 0} {e^(x+h)-e^(x)}/{h} #

# " " = lim_{h to 0} {e^xe^h-e^(x)}/{h} #

# " " = lim_{h to 0} e^x((e^h-1))/{h} #

# " " = e^xlim_{h to 0} ((e^h-1))/{h} #

Think about this limit for a moment and we can rewrite it as:

#lim_{h to 0} ((e^h-1))/{h} = lim_{h to 0} ((e^h-e^0))/{h} #

# " " = lim_{h to 0} ((e^(0+h)-e^0))/{h} #

# " " = f'(0) # (by the derivative definition)

Hence,

# f'(x) = e^xf'(0) #

Now, It can be shown that this limit:

# f'(0) = lim_{h to 0} ((e^h-1))/{h} #

both exists and is equal to unity. Additionly, the number

And so:

# d/dx e^x=e^x#

This special exponential function with Euler's number, **only function** that remains unchanged when differentiated.

**Method 2 - Power Series**

We can use the power series:

# e^x = 1 +x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ... #

Then we can differentiate term by term using the power rule:

# d/dx e^x = d/dx{1 +x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ... } #

# \ \ \ \ \ \ \ \ \ = 0 +1 + (2x)/(2!) + (3x^2)/(3!) + (4x^3)/(4!) + (5x^4)/(5!) + ... #

# \ \ \ \ \ \ \ \ \ = 1 + (x)/(1!) + (3x^2)/(2! * 2) + (4x^3)/(3! * 4) + (5x^4)/(4! * 5) + ... #

# \ \ \ \ \ \ \ \ \ = 1 +x + x^2/(2!) + x^3/(3!) + x^4/(4!) + ... #