Server Error

Server Not Reachable.

This may be due to your internet connection or the nubtrek server is offline.

Thought-Process to Discover Knowledge

Welcome to nubtrek.

Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.

In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators.

Read in the blogs more about the unique learning experience at nubtrek.continue
mathsDifferential CalculusStandard Results in Derivatives

Derivatives of Exponents and Logarithmic Functions

In this page, derivatives of exponents and logarithmic functions such as `e^x`, `a^x`, and `ln x`

click on the content to continue..

Finding the derivative of `y=e^(x)` in first principles:

`d/(dx) e^(x)`

`=lim_(delta->0) [color(coral)(e^(x+delta)``- color(deepskyblue)(e^x)]//delta`

`=lim_(delta->0) [color(coral)(e^x xx e^(delta))``- color(deepskyblue)(e^x)]//delta`

`=lim_(delta->0) e^x[color(coral)( e^(delta))``- color(deepskyblue)(1)]//delta`

applying the standard limit `lim_(p->0)(e^p - 1)//p = 1`

`= e^(x)`

What does the above prove?

  • `d/(dx) e^(x) = e^(x)`
  • `d/(dx) e^(x) = e^(x)`
  • `d/(dx) e^(ax) = e^(ax)`

The answer is "`d/(dx) e^(x) = e^(x)`".

Finding the derivative of `y=ln x` :

`y=ln x`


differentiating the equation

applying chain rule `(d)/(dy)e^y (dy)/(dx)=1`

`e^y (dy)/(dx)=1`

`x (dy)/(dx)=1`


What does the above prove?

  • `d/(dx) ln x = 1/x`
  • `d/(dx) ln x = 1/x`
  • `d/(dx) ln x = cos x + sin x`

The answer is "`d/(dx) ln x = 1/x`"

Finding the derivative of `y=a^x` :

substituting `a=e^(ln a)`

`y=e^(x ln a )`

differentiating the equation
`(dy)/(dx)=(d)/(dx) e^(x ln a )`

applying chain rule with `u=x ln a`
`(dy)/(dx)=(d)/(du) e^u (d)/(dx)(x ln a )`

`(dy)/(dx)= e^u xx ln a`

`(dy)/(dx)= e^(x ln a ) xx ln a `

substituting `e^(ln a)=a`

`(dy)/(dx)=a^x ln a `

What does the above prove?

  • `d/(dx) a^x = a^x ln a`
  • `d/(dx) a^x = a^x ln a`
  • `d/(dx) a^x = a^x`

The answer is "`d/(dx) a^x = a^x ln a`"

Derivatives of Exponents or Logarithmic Functions:
`d/(dx) e^x = e^x`

`d/(dx) a^x = a^x ln a`

`d/(dx) ln x = 1/x `

  `d/(dx) e^x = e^x`
definition of `e` is rate of change is proportional to itself

  `d/(dx) a^x = a^x ln a`
`a` equals `e^(lna)`

  `d/(dx) ln x = 1/x `
natural log is inverse of `e` power

Solved Exercise Problem:

What is the derivative of `log_(10) x`?
Note: Use the identity `log_(10) x = (log_e x)/(log_e 10)`

  • `1/(10x)`
  • `1/(xln 10)`
  • `1/(xln 10)`

The answer is "`1/(xln 10)`".
` d/(dx)log_10 x`
`= d/(dx)(log_e x)/(log_e 10)`
`= 1/(log_e 10) d/(dx) log_e x`
`=1/(log_e 10) xx 1/x `
`=1/(xlog_e 10)`

Solved Exercise Problem:

What is the derivative of `e^(ax)`?

  • `ae^(ax)`
  • `ae^(ax)`
  • `e^(ax)`

The answer is "`ae^(ax)`". Applying chain rule with `u=ax`,
`d/(du) e^u d/(dx) ax`
`e^u xx a`

slide-show version coming soon