$e^{lnx} = x$

ln$e^x = x$

If $f(x) = e^x$, $f'(x) = e^x$

If $f(x) = e^{g(x)}$, $f'(x) = g'(x)e^{g(x)}$

If $f(x) = $ln$x$, $f'(x) = 1/x$

If $f(x) = a^x$, $f'(x) = ($ln$a)a^x$

For $f(x) = a^x$, $f'(x)$ at $x=0$ is ln$a$.

If $f(x) = $log$_{a}x$, $f'(x) = 1/{xlna}$

If $f(x) = $ln$g(x)$, $f'(x) = {g'(x)}/{g(x)}$

Example: find the derivative of $f(x) = e^{3x^2}$:

$f'(x) = g'(x)e^{g(x)} = 6xe^{3x^2}$

Example: find the derivative of $f(x) = $ln$(3x^2)$: $f'(x) = {g'(x)}/{g(x)} = {6x}/{3x^2} = 2/x$

The letter $e$ was selected to represent the irrational number 2.718281828459, in honour of the great Swiss mathematician Leonhard Euler (1706 - 1781).

$e$ is defined as the limit of $(1 + 1/n)^n$ as n approaches ∞.

It may also be expressed as:

$$e = ∑↙{n=0}↖{∞} 1/{n!} = 1 + 1/1 + 1/{1⋅2} + 1/{1⋅2⋅3} + ...$$Observing the graph of sine of $x$, it can be seen that the slope is graphed by cosine $x$. When $x$ is zero, sine is zero, with increasing tangent gradient. At zero, the slope is maximum, and gradually decreases, but not constantly, to where it levels out at $π/2$, where the tangent gradient is zero.

cosine is therefore the first derivative of the sine function.

Since the sine and cosine graphs superimpose if sine is translated $π/2$ to the left, cos$x = $sin$(x+π/2)$.

$d/{dx}($cos$x) = d/{dx}[$sin$(x+ π/2)] $

$= [$cos$(x+π/2)]⋅1 = $cos$(x+π/2)$

cos$(x+π/2) = -$sin$x$.

The derivative of tan$x$ can also be derived by a similar procedure:

$d/{dx}($tan$x) = d/{dx}({sinx}/{cosx}) = {cosx(cosx) - sinx(-sinx)}/{(cosx)^2} = {cos^2x+sin^2x}/{cos^2x} = 1/{cos^2x}$, cos$x ≠0$

If $f(x) = $sin$ x$, then $f'(x) = $cos$ x$

If $f(x) = $cos$ x$, then $f'(x) = -$sin$ x$

If $f(x) = $tan$ x$, then

$f'(x) = $se$c^2 x$ $ ( = 1/{cos^2x}) = 1 + $tan$^2x$

If $f(x) = $arcsin$ x$, then $f'(x) = 1/{√{1-x^2}}$

If $f(x) = $arccos$ x$, then $f'(x) = -1/{√{1-x^2}}$

If $f(x) = $arctan$ x$, then $f'(x) = 1/{1+x^2}$

If $f(x) = $arccot$ x$, then $f'(x) = -1/{1+x^2}$

If $f(x) = $sin$ {x/2}$, then $f'(x) = {1/2}$cos$ {x/2}$

If $f(x) = $cos$ 3x$, then $f'(x) = -3$sin$ 3x$

If $f(x) = $sin$(2x - 1)$, then $f'(x) = 2$cos$(2x - 1)$

If $f(x) = $cot$x$, then

$f'(x) = -$csc$^2x = -1/{sin^2(x)} = -1 - $cot$^2(x) $

Content © Renewable-Media.com. All rights reserved. Created : January 23, 2015

The most recent article is:

View this item in the topic:

and many more articles in the subject:

Physics is the science of the very small and the very large. Learn about Isaac Newton, who gave us the laws of motion and optics, and Albert Einstein, who explained the relativity of all things, as well as catch up on all the latest news about Physics, on ScienceLibrary.info.

1710 - 1790

Johann Bernoulli (II) was the son of Johan (I), and the younger brother of Daniel and Nicolaus, all noted mathematicians.

Website © renewable-media.com | Designed by: Andrew Bone