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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

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» Derivatives of Exponents or Logarithmic Functions

`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*

*plain and simple summary*

nub

*plain and simple summary*

nub

dummy

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 `

*simple steps to build the foundation*

trek

*simple steps to build the foundation*

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

starting on learning "Derivatives of Exponents and Logarithmic Functions". In this page, derivatives of exponents and logarithmic functions such as e power x, a power x, and natural log of x.

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^(ax) = e^(ax)`

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

Finding the derivative of `y=ln x` :

`y=ln x`

`e^y=x`*differentiating the equation*

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

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

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

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

What does the above prove?

- `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`

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

*comprehensive information for quick review*

Jogger

*comprehensive information for quick review*

Jogger

dummy

`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*

*practice questions to master the knowledge*

Exercise

*practice questions to master the knowledge*

Exercise

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)`

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)`

What is the derivative of `e^(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`

`ae^(ax)`

*your progress details*

Progress

*About you*

Progress

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)

=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

= e^(x)

What does the above prove?

1

2

The answer is "d by dx of e power x = e power x"

Finding the derivative of y = natural log x is given. What does the above prove.

1

2

The answer is "d by dx of natural log x = 1 by x "

Finding the derivative of y = a power x is given. What does the above prove.

1

2

The answer is "d by dx of a power x = a power x, multiplied, natural log a"

Derivatives of Exponents or Logarithmic Functions are listed

Derivatives of Exponents or Logarithmic Functions are given. The number e is defined such that rate of change is proportional to itself. All follow a certain pattern, quickly follow them to derive the result.

what is the derivative of log base 10 x.

1

2

The answer is "1 by x natural log 10"

What is the derivative of e power a x .

1

2

The answer is "a e power a x"