5.13 More DerivativesHL ONLY

1. Derivatives of Exponential and Logarithmic Functions

We already know that the derivative of the natural exponential function $e^x$ is itself, and the derivative of $\ln x$ is $\dfrac{1}{x}$. For general bases $a > 0$ (where $a \neq 1$), the differentiation rules are expanded as follows:

Function $f(x)$ Derivative $f'(x)$
$a^x$ $a^x \ln a$
$\log_a x$ $\dfrac{1}{x \ln a}$

Note: These formulas can be directly derived using the change of base formulas: $a^x = e^{x \ln a}$ and $\log_a x = \dfrac{\ln x}{\ln a}$.

EXAMPLE 1

Find the derivative of $f(x) = 3^x + x^3$.

Solution:
Applying the power rule to $x^3$ and the general exponential rule to $3^x$ $$ f'(x) = \mathbf{3^x \ln 3 + 3x^2} $$

EXAMPLE 2

Find the derivative of $y = \log_5(x^2 + 4)$.

Solution:
We use the chain rule. Let $u = x^2 + 4$, then $y = \log_5(u)$.
The derivative of the outer function is $\dfrac{1}{u \ln 5}$, and the derivative of the inner function is $2x$.
$$ \dfrac{dy}{dx} = \dfrac{1}{(x^2+4) \ln 5} \cdot (2x) = \mathbf{\dfrac{2x}{(x^2+4) \ln 5}} $$

2. Derivatives of Inverse Trigonometric Functions

The derivatives of the inverse trigonometric functions ($\arcsin x$, $\arccos x$, and $\arctan x$) are standard results in the HL syllabus. They are essential for both differential calculus and integration.

Function $f(x)$ Derivative $f'(x)$
$\arcsin x$ $\dfrac{1}{\sqrt{1-x^2}}$
$\arccos x$ $-\dfrac{1}{\sqrt{1-x^2}}$
$\arctan x$ $\dfrac{1}{1+x^2}$

Graphical Representation

Below are the precise graphs of $y = \arcsin x$ and $y = \arctan x$. Notice the restricted domains and ranges that make these inverse relations valid functions.

x y 1 -1 π/2 -π/2 y = arcsin x
x y y = π/2 y = -π/2 y = arctan x

EXAMPLE 3

Differentiate the function $y = \arcsin(3x)$.

Solution:
Utilizing the chain rule where the inner function is $3x$:
$$ y' = \dfrac{1}{\sqrt{1-(3x)^2}} \cdot (3x)' = \dfrac{1}{\sqrt{1-9x^2}} \cdot 3 = \mathbf{\dfrac{3}{\sqrt{1-9x^2}}} $$

EXAMPLE 4

Find the derivative of $f(x) = x^2 \arctan(x)$.

Solution:
Here, we must apply the Product Rule: $(uv)' = u'v + uv'$. Let $u = x^2$ and $v = \arctan x$.
$u' = 2x$ and $v' = \dfrac{1}{1+x^2}$.
$$ \begin{aligned} f'(x) &= (2x)(\arctan x) + (x^2)\left(\dfrac{1}{1+x^2}\right) \\ &= \mathbf{2x \arctan x + \dfrac{x^2}{1+x^2}} \end{aligned} $$

3. Monotonicity of Inverse Trigonometric Functions

We can use the derivative to rigorously determine the monotonicity of these functions:

  • Since $\dfrac{1}{\sqrt{1-x^2}} > 0$ for all $x \in (-1, 1)$, $y = \arcsin x$ is strictly increasing on its entire domain.
  • Since $-\dfrac{1}{\sqrt{1-x^2}} < 0$ for all $x \in (-1, 1)$, $y = \arccos x$ is strictly decreasing on its entire domain.
  • Since $\dfrac{1}{1+x^2} > 0$ for all $x \in \mathbb{R}$, $y = \arctan x$ is strictly increasing on its entire domain.

EXAMPLE 5

Determine if the function $g(x) = \arccos(2x)$ is increasing or decreasing.

Solution:
We find the first derivative using the chain rule:
$$ g'(x) = -\dfrac{1}{\sqrt{1-(2x)^2}} \cdot (2) = -\dfrac{2}{\sqrt{1-4x^2}} $$ Because the square root is always positive, the fraction $\dfrac{2}{\sqrt{1-4x^2}}$ is positive. With the negative sign in front, we have $g'(x) < 0$ for all $x \in \left(-\dfrac{1}{2}, \dfrac{1}{2}\right)$.
Hence, $g(x)$ is strictly decreasing on its domain.