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Derivatives of trigs, exponents and logs

Logs and Exponents

Natural log
Natural log and its first differential 1/x

$e^{lnx} = x$

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)$: $f'(x) = {g'(x)}/{g(x)} = {6x}/{3x^2} = 2/x$

e exponent x is its own derivative

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:

$$e = ∑↙{n=0}↖{∞} 1/{n!} = 1 + 1/1 + 1/{1⋅2} + 1/{1⋅2⋅3} + ...$$

Derivatives of Trigonometric Functions

Sine and cosine

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 $f(x) = sinx$ from first principles:
$f'(x) = {lim}↙{h→0} {f(x+h) - f(x)}/h$
     $= {lim}↙{h→0} {sin(x+h) - sin(x)}/h$
     $= {lim}↙{h→0} ({sin x⋅cos h + cos x ⋅ sin h - sin x}/h)$
     $= {lim}↙{h→0} ({cos x⋅ {sinh}/h + sinx⋅ {cosh -1}/h)$
     $= cosx⋅{lim}↙{h→0} {sinh}/h + sinx⋅{lim}↙{h→0} {cosh - 1}/h$
     $= cosx⋅1 + sinx⋅0 = cosx$

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$x ≠0$

Basic Trig Differentials

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$

Basic Trig Differentials

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 $f'(x) = -1/{1+x^2}$

further Trig Differentials

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 © Andrew Bone. All rights reserved. Created : January 23, 2015

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