 # Derivatives of trigs, exponents and logs

## Logs and Exponents 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\$ 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 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) \$

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