Differentiation and Derivatives

The Definition of Derivative

The derivative of a function $f(x)$ with respect to $x$ , denoted as $f'(x)$ , is defined as:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$
This definition is also known as the first principle of differentiation.

More Details:

Purpose: The derivative $f'(x)$ represents the instantaneous rate of change of the function $f(x)$ at a particular point.
Interpretation: Geometrically, it gives the slope of the tangent line to the graph of $f(x)$ at that point.
Limit Existence: For the derivative to exist at a point, the limit must exist. This means the left-hand limit and the right-hand limit must be equal.
Differentiability: A function is said to be differentiable at a point if its derivative exists at that point. If a function is differentiable at all points in its domain, it is said to be a differentiable function.

Example:

Find $f'(x)$ of $f(x) = 3x^2 - 2$ using the definition of derivative.

Step 1: Write down the formula of the definition of derivative
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Step 2: Determine $f(x+h)$
$f(x+h) = 3(x+h)^2 - 2 = 3(x^2 + 2xh + h^2) - 2 = 3x^2 + 6xh + 3h^2 - 2$

Step 3: Substitute into the formula and simplify the limit
$f'(x) = \lim{h \to 0} \frac{(3x^2 + 6xh + 3h^2 - 2) - (3x^2 - 2)}{h} = \lim{h \to 0} \frac{6xh + 3h^2}{h} = \lim_{h \to 0} (6x + 3h) = 6x$

Step 4: Write the final answer: The derivative of $f(x)$ is $f'(x) = 6x$ .

Examples of Differentiation Using the Definition

Polynomial

(a) $f(x) = x^2 - 10x + 3$

Step 1:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Step 2:
$f(x+h) = (x+h)^2 - 10(x+h) + 3 = x^2 + 2xh + h^2 - 10x - 10h + 3$

Step 3:
\begin{aligned}
f'(x) &= \lim{h \to 0} \frac{(x^2 + 2xh + h^2 - 10x - 10h + 3) - (x^2 - 10x + 3)}{h} \ &= \lim{h \to 0} \frac{2xh + h^2 - 10h}{h} \
&= \lim_{h \to 0} (2x + h - 10) \
&= 2x - 10
\end{aligned}

Step 4:
$f'(x) = 2x - 10$

Radical

(b) $f(x) = \sqrt{x} - 1$

Step 1:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Step 2:
$f(x+h) = \sqrt{x+h} - 1$

Step 3:
\begin{aligned}
f'(x) &= \lim{h \to 0} \frac{(\sqrt{x+h} - 1) - (\sqrt{x} - 1)}{h} \ &= \lim{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \
&= \lim{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})} \ &= \lim{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})} \
&= \lim{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})} \ &= \lim{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} \
&= \frac{1}{2\sqrt{x}}
\end{aligned}

Step 4:
$f'(x) = \frac{1}{2\sqrt{x}}$

Rational

Step 1:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Step 2:
$f(x+h) = \frac{6}{(x+h)+1} = \frac{6}{x+h+1}$

Step 3:
\begin{aligned}
f'(x) &= \lim{h \to 0} \frac{\frac{6}{x+h+1} - \frac{6}{x+1}}{h} \ &= \lim{h \to 0} \frac{\frac{6(x+1) - 6(x+h+1)}{(x+h+1)(x+1)}}{h} \
&= \lim{h \to 0} \frac{6x + 6 - 6x - 6h - 6}{h(x+h+1)(x+1)} \ &= \lim{h \to 0} \frac{-6h}{h(x+h+1)(x+1)} \
&= \lim_{h \to 0} \frac{-6}{(x+h+1)(x+1)} \
&= \frac{-6}{(x+1)(x+1)} \
&= \frac{-6}{(x+1)^2}
\end{aligned}

Step 4:
$f'(x) = \frac{-6}{(x+1)^2}$

Rational + Radical

(d) $f(x) = \frac{1}{\sqrt{x} - 4}$

Step 1:
$f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Step 2:
$f(x+h) = \frac{1}{\sqrt{x+h} - 4}$

Step 3:
\begin{aligned}
f'(x) &= \lim{h \to 0} \frac{\frac{1}{\sqrt{x+h} - 4} - \frac{1}{\sqrt{x} - 4}}{h} \ &= \lim{h \to 0} \frac{\frac{(\sqrt{x} - 4) - (\sqrt{x+h} - 4)}{(\sqrt{x+h} - 4)(\sqrt{x} - 4)}}{h} \
&= \lim{h \to 0} \frac{\sqrt{x} - \sqrt{x+h}}{h(\sqrt{x+h} - 4)(\sqrt{x} - 4)} \ &= \lim{h \to 0} \frac{(\sqrt{x} - \sqrt{x+h})(\sqrt{x} + \sqrt{x+h})}{h(\sqrt{x+h} - 4)(\sqrt{x} - 4)(\sqrt{x} + \sqrt{x+h})} \
&= \lim{h \to 0} \frac{x - (x+h)}{h(\sqrt{x+h} - 4)(\sqrt{x} - 4)(\sqrt{x} + \sqrt{x+h})} \ &= \lim{h \to 0} \frac{-h}{h(\sqrt{x+h} - 4)(\sqrt{x} - 4)(\sqrt{x} + \sqrt{x+h})} \
&= \lim_{h \to 0} \frac{-1}{(\sqrt{x+h} - 4)(\sqrt{x} - 4)(\sqrt{x} + \sqrt{x+h})} \
&= \frac{-1}{(\sqrt{x} - 4)(\sqrt{x} - 4)(\sqrt{x} + \sqrt{x})} \
&= \frac{-1}{2\sqrt{x}(\sqrt{x} - 4)^2}
\end{aligned}

Step 4:
$f'(x) = \frac{-1}{2\sqrt{x}(\sqrt{x} - 4)^2}$

Rules of Derivative / Differentiation

For $y = f(x)$ , the notation of the derivative is:
$y' = f'(x) = \frac{dy}{dx} = \frac{d}{dx}f(x)$

Constant Rule

Let $y = f(x) = a$ , where $a$ is a constant, then
$f'(x) = \frac{dy}{dx} = 0$

Power Rule

Let $y = f(x) = ax^n$ , then
$\frac{dy}{dx} = f'(x) = nax^{n-1}$

Example 2.2
Differentiate the following with respect to x.
a) $y = 10$
$\frac{dy}{dx} = 0$

b) $y = -x^5 + x^8 - 12$
$\frac{dy}{dx} = -5x^4 + 8x^7$

c) $y = \sqrt{x} = x^{\frac{1}{2}}$
$\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}$

d) $y = -\frac{2}{3\sqrt{x}} = -\frac{2}{3}x^{-\frac{1}{2}}$
$\frac{dy}{dx} = -\frac{2}{3} \cdot \left(-\frac{1}{2}\right) x^{-\frac{3}{2}} = \frac{1}{3x^{\frac{3}{2}}}$

Product Rule

If $y = f(x) \cdot g(x)$ , then let $u = f(x)$ and $v = g(x)$
$\frac{dy}{dx} = u'v + uv'$

Example 2.3
Differentiate the following with respect to x.

(a) $y = (3x + 11)(2x - 7)$
Let $u = 3x + 11$ , $v = 2x - 7$
$u' = 3$ , $v' = 2$
$\frac{dy}{dx} = (3)(2x - 7) + (3x + 11)(2) = 6x - 21 + 6x + 22 = 12x + 1$

Alternative: Expand the $y$ .
$y = 6x^2 - 21x + 22x - 77 = 6x^2 + x - 77$
$\frac{dy}{dx} = 12x + 1$

(b) $f(x) = (x^2 + 1)(x - 1)$
Let $u = x^2 + 1$ , $v = x - 1$
$u' = 2x$ , $v' = 1$
$f'(x) = (2x)(x - 1) + (x^2 + 1)(1) = 2x^2 - 2x + x^2 + 1 = 3x^2 - 2x + 1$

Alternative: Expand the $f(x)$ .
$f(x) = x^3 - x^2 + x - 1$
$f'(x) = 3x^2 - 2x + 1$

Quotient Rule

If $y = \frac{f(x)}{g(x)}$ , then let $u = f(x)$ and $v = g(x)$
$\frac{dy}{dx} = \frac{vu' - uv'}{v^2}$

Example 2.4
Differentiate the following with respect to x.

(a) $y = \frac{2x + 3}{3x - 5}$
Let $u = 2x + 3$ , $v = 3x - 5$
$u' = 2$ , $v' = 3$
$\frac{dy}{dx} = \frac{(3x - 5)(2) - (2x + 3)(3)}{(3x - 5)^2} = \frac{6x - 10 - 6x - 9}{(3x - 5)^2} = \frac{-19}{(3x - 5)^2}$

(b) $f(x) = \frac{5x^2 + 11}{x - 4}$
Let $u = 5x^2 + 11$ , $v = x - 4$
$u' = 10x$ , $v' = 1$
$\frac{dy}{dx} = \frac{(x - 4)(10x) - (5x^2 + 11)(1)}{(x - 4)^2} = \frac{10x^2 - 40x - 5x^2 - 11}{(x - 4)^2} = \frac{5x^2 - 40x - 11}{(x - 4)^2}$

Derivative of Trigonometric Functions

$y = \sin(x)$ , $\frac{dy}{dx} = \cos(x)$
$y = \cos(x)$ , $\frac{dy}{dx} = -\sin(x)$
$y = \tan(x)$ , $\frac{dy}{dx} = \sec^2(x)$
$y = \sec(x)$ , $\frac{dy}{dx} = \sec(x)\tan(x)$
$y = \csc(x)$ , $\frac{dy}{dx} = -\csc(x)\cot(x)$
$y = \cot(x)$ , $\frac{dy}{dx} = -\csc^2(x)$

Example 2.5
Find the derivative of the following functions.
a) $y = 2\ln(x) - \sin(x) + x e^3$
$\frac{dy}{dx} = \frac{2}{x} - \cos(x) + e^3$

b) $y = \frac{1}{x} - 3\sec(x) - \ln(6) + 5x$
$\frac{dy}{dx} = -\frac{1}{x^2} - 3\sec(x)\tan(x) + 5$

c) $y = \ln(x)\tan(x)$
Let $u = \ln(x)$ , $v = \tan(x)$
$u' = \frac{1}{x}$ , $v' = \sec^2(x)$
$\frac{dy}{dx} = \frac{1}{x}\tan(x) + \ln(x)\sec^2(x)$

d) $y = \frac{e^x}{\cos(x)}$
Let $u = e^x$ , $v = \cos(x)$
$u' = e^x$ , $v' = -\sin(x)$
$\frac{dy}{dx} = \frac{\cos(x)e^x - e^x(-\sin(x))}{\cos^2(x)} = \frac{e^x(\cos(x) + \sin(x))}{\cos^2(x)}$

Chain Rule

If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$ , then the composite function $F(x) = f(g(x))$ has the derivative:
$F'(x) = f'(g(x)) \cdot g'(x)$
In Leibniz notation, if $y = f(u)$ and $u = g(x)$ , then
$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Alternative using Chain Rule:
Let $y = u^2$ where $u = x^2 - 3$
$\frac{dy}{du} = 2u$ and $\frac{du}{dx} = 2x$
$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = (2u)(2x) = 2(x^2 - 3)(2x) = 4x(x^2 - 3)$

b) $F(x) = \sqrt[9]{x - 1} = (x - 1)^{\frac{1}{9}}$
Composite Function $F(x) = f(g(x))$
Let $f(u) = u^{\frac{1}{9}}$ and $g(x) = x - 1$
$f'(u) = \frac{1}{9}u^{-\frac{8}{9}}$ and $g'(x) = 1$
$F'(x) = f'(g(x)) \cdot g'(x) = \frac{1}{9}(x - 1)^{-\frac{8}{9}}(1) = \frac{1}{9(x - 1)^{\frac{8}{9}}}$

Alternative using Chain Rule:
Let $y = u^{\frac{1}{9}}$ where $u = x - 1$
$\frac{dy}{du} = \frac{1}{9}u^{-\frac{8}{9}}$ and $\frac{du}{dx} = 1$
$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{9}u^{-\frac{8}{9}}(1) = \frac{1}{9(x - 1)^{\frac{8}{9}}}$

Note: By using the chain rule method, the working step is from the outside to the inside.
$\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$
With the combination of the Power Rule and the Chain Rule, the derivative process is as below:
If $y = [u]^n$ , then
$\frac{dy}{dx} = n[u]^{n-1} \cdot f'(u)$
If $y = [f(x)]^n$ , then
$\frac{dy}{dx} = n[f(x)]^{n-1} \cdot f'(x)$
If $y = [f(g(x))]^n$ , then
$\frac{dy}{dx} = n[f(g(x))]^{n-1} \cdot f'(g(x)) \cdot g'(x)$

Polynomial, Rational, Radical Functions

Example 2.7
Differentiate the following with respect to x.
a) $y = \sqrt{4x^3}$
$\frac{dy}{dx} = \frac{3}{2}(4x^3)^{-\frac{1}{2}} \cdot (12x^2) = \frac{18x^2}{\sqrt{4x^3}}$

b) $y = \frac{5}{\sqrt[9]{x^4}} = 5x^{-\frac{4}{9}}$
$\frac{dy}{dx} = -\frac{20}{9}x^{-\frac{13}{9}} = -\frac{20}{9\sqrt[9]{x^{13}}}$

c) $y = (x^2 + 10)^{3}$
$\frac{dy}{dx} = 3(x^2 + 10)^{2} \cdot (2x) = 6x(x^2 + 10)^2$

d) $y = (x + 1)^3 (x - 1)$
Product Rule: $\frac{dy}{dx} = u'v + uv'$
Let $u = (x+1)^3$ , $v = x-1$
$u' = 3(x+1)^2$ , $v' = 1$
$\frac{dy}{dx} = 3(x+1)^2 (x-1) + (x+1)^3 (1) = (x+1)^2[3(x-1) + (x+1)] = (x+1)^2[3x - 3 + x + 1] = (x+1)^2(4x - 2) = 2(x+1)^2(2x - 1)$

Trigonometry Functions

Example 2.8
Differentiate the following with respect to x.
a) $y = \sin(5x)$
$\frac{dy}{dx} = \cos(5x) \cdot 5 = 5\cos(5x)$

b) $y = \cos(\sin(x))$
$\frac{dy}{dx} = -\sin(\sin(x)) \cdot \cos(x) = -\cos(x)\sin(\sin(x))$

c) $y = \tan^6(3x)$
$\frac{dy}{dx} = 6\tan^5(3x) \cdot \sec^2(3x) \cdot 3 = 18\sec^2(3x)\tan^5(3x)$

d) $f(x) = \tan \left[\frac{2 - x^2}{2 + x^2}\right]$
\begin{aligned}
f'(x) &= \sec^2 \left[\frac{2 - x^2}{2 + x^2}\right] \cdot \left[\frac{(2 + x^2)(-2x) - (2 - x^2)(2x)}{(2 + x^2)^2}\right] \
&= \sec^2 \left[\frac{2 - x^2}{2 + x^2}\right] \cdot \left[\frac{-4x - 2x^3 - 4x + 2x^3}{(2 + x^2)^2}\right] \
&= \sec^2 \left[\frac{2 - x^2}{2 + x^2}\right] \cdot \left[\frac{-8x}{(2 + x^2)^2}\right]
\end{aligned}

Exponential and Logarithmic Functions

Example 2.9
Differentiate the following with respect to x.
a) $y = e^{\sin x}$
$\frac{dy}{dx} = e^{\sin x} \cdot \cos x = \cos x \cdot e^{\sin x}$

b) $y = 5e^x \cdot x^3$
$\frac{dy}{dx} = 5(e^x \cdot 3x^2 + e^x \cdot x^3) = 5x^2e^x(3+x)$

c) $y = \ln(cos x)$
$\frac{dy}{dx} = -\tan(x)$

d) $y = \ln \left( \frac{2 - x}{2 + x} \right)$
\begin{aligned}
\frac{dy}{dx} &= \frac{1}{\frac{2 - x}{2 + x}} \cdot \frac{(-1)(2 + x) - (1)(2 - x)}{(2 + x)^2} \
&= \frac{2 + x}{2 - x} \cdot \frac{-2 - x -2 + x}{(2 + x)^2} \
&= \frac{2 + x}{2 - x} \cdot \frac{-4}{(2 + x)^2} \
&= \frac{-4}{(2 - x)(2 + x)} \
&= \frac{-4}{4 - x^2}
\end{aligned}

Higher Order of Differentiation

Let $y = f(x)$

The first (1st) derivative, $\frac{dy}{dx} = f'(x)$
The second (2nd) derivative, $\frac{d^2y}{dx^2} = f''(x)$
The third (3rd) derivative, $\frac{d^3y}{dx^3} = f'''(x)$

Example 2.10
Find the second derivative for the following.
a) $y = -x^5 + 3x^4 + x^2$
$\frac{dy}{dx} = -5x^4 + 12x^3 + 2x$
$\frac{d^2y}{dx^2} = -20x^3 + 36x^2 + 2$

b) $f(x) = e^{3x} - \sin(4x)$
$f'(x) = 3e^{3x} - 4\cos(4x)$
$f''(x) = 9e^{3x} + 16\sin(4x)$

Implicit Differentiation

Previously, all the functions we met can be expressed a variable explicitly to another variable. For instance, $y = x^3 - x - 4$ or $e^x = 0$ , in general way is $y = f(x)$ . However, some function is expressed implicitly between the variable and y. Likely, $x = 0$ . It is quite difficult to rearrange y in terms of x. In order to find the first derivative of this kind of equation, so we need to differentiate y implicitly with respect to x.

Solving steps for implicit differentiation

Step 1: Differentiate both sides (every term) of the equation with respect to x
Step 2: Rearrange the terms with on LHS, the others on RHS
Step 3: Factorize and simplify.

Example 2.11

Find for $cos(x) = 2$ by using implicit differentiation.
*Step 1: Differentiate every terms w.r.t. x. ( 𝑚𝑑𝑦𝑑𝑥 𝑜𝑛 𝐿𝐻𝑆, 𝑜𝑡ℎ𝑒𝑟𝑠 𝑜𝑛 𝑅𝐻𝑆 (( 𝑚𝑑𝑦𝑑𝑥 𝑎𝑛𝑑 𝑆𝑖𝑚𝑝𝑙𝑖𝑓𝑦 () 𝑆𝑡𝑒𝑝 3: 𝐹𝑎𝑐𝑡𝑜𝑟𝑖𝑧𝑒
Find for by using implicit differentiation.
*Step 1: Differentiate every terms w.r.t. x.

Implicit Differentiation Table

Following are the implicit differentiation of common function:

(1) Power Rule

For $+ 3 = 2 𝑡𝑎𝑛 ((𝑥𝑦)$ :

*   Step 1: Differentiate every terms w.r.t. x. 2 𝑦 2 𝑠𝑒𝑐 𝑚𝑑𝑦𝑑𝑥+ 𝑥𝑦 𝑠𝑒𝑐 𝑚𝑑𝑦𝑑𝑥=3++2 , 2 2 = + .

_Step 2: Rearrange 𝑚𝑑𝑦𝑑𝑥 on 𝐿

The Definition of Derivative The derivative of a function $f(x)$ with respect to $x$ , denoted as $f'(x)$ , is defined as: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ This definition is also known as the first principle of differentiation.

More Details: - Purpose: The derivative $f'(x)$ represents the instantaneous rate of change of the function $f(x)$ at a particular point. - Interpretation: Geometrically, it gives the slope of the tangent line to the graph of $f(x)$ at that point. This tangent line visually demonstrates how the function is changing at an infinitesimally small interval around that point. - Limit Existence: For the derivative to exist at a point, the limit must exist. This means the left-hand limit and the right-hand limit must be equal, ensuring that the function approaches the same value from both sides. - Differentiability: A function is said to be differentiable at a point if its derivative exists at that point. If a function is differentiable at all points in its domain, it is said to be a differentiable function. Differentiability implies continuity, but the converse is not always true (e.g., a function with a sharp corner is continuous but not differentiable at that corner).

Example: Find $f'(x)$ of $f(x) = 3x^2 - 2$ using the definition of derivative.

Step 1: Write down the formula of the definition of derivative $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Step 2: Determine $f(x+h)$ $f(x+h) = 3(x+h)^2 - 2 = 3(x^2 + 2xh + h^2) - 2 = 3x^2 + 6xh + 3h^2 - 2$

Step 3: Substitute into the formula and simplify the limit $f'(x) = \lim{h \to 0} \frac{(3x^2 + 6xh + 3h^2 - 2) - (3x^2 - 2)}{h} = \lim{h \to 0} \frac{6xh + 3h^2}{h} = \lim_{h \to 0} (6x + 3h) = 6x$

Step 4: Write the final answer: The derivative of $f(x)$ is $f'(x) = 6x$ . ### Examples of Differentiation Using the Definition #### Polynomial (a) $f(x) = x^2 - 10x + 3$

Step 1: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Step 2: $f(x+h) = (x+h)^2 - 10(x+h) + 3 = x^2 + 2xh + h^2 - 10x - 10h + 3$

Step 3: $\begin{aligned} f'(x) &= \lim{h \to 0} \frac{(x^2 + 2xh + h^2 - 10x - 10h + 3) - (x^2 - 10x + 3)}{h} \ &= \lim{h \to 0} \frac{2xh + h^2 - 10h}{h} \ &= \lim_{h \to 0} (2x + h - 10) \ &= 2x - 10 \ \end{aligned}$

Step 4: $f'(x) = 2x - 10$ #### Radical (b) $f(x) = \sqrt{x} - 1$

Step 1: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Step 2: $f(x+h) = \sqrt{x+h} - 1$

Step 3: $\begin{aligned} f'(x) &= \lim{h \to 0} \frac{(\sqrt{x+h} - 1) - (\sqrt{x} - 1)}{h} \ &= \lim{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} \ &= \lim{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})} \ &= \lim{h \to 0} \frac{(x+h) - x}{h(\sqrt{x+h} + \sqrt{x})} \ &= \lim{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})} \ &= \lim{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} \ &= \frac{1}{2\sqrt{x}} \ \end{aligned}$

Step 4: $f'(x) = \frac{1}{2\sqrt{x}}$ #### Rational (c) $f(x) = \frac{6}{x+1}$

Step 1: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Step 2: $f(x+h) = \frac{6}{(x+h)+1} = \frac{6}{x+h+1}$

Step 3: $\begin{aligned} f'(x) &= \lim{h \to 0} \frac{\frac{6}{x+h+1} - \frac{6}{x+1}}{h} \ &= \lim{h \to 0} \frac{\frac{6(x+1) - 6(x+h+1)}{(x+h+1)(x+1)}}{h} \ &= \lim{h \to 0} \frac{6x + 6 - 6x - 6h - 6}{h(x+h+1)(x+1)} \ &= \lim{h \to 0} \frac{-6h}{h(x+h+1)(x+1)} \ &= \lim_{h \to 0} \frac{-6}{(x+h+1)(x+1)} \ &= \frac{-6}{(x+1)(x+1)} \ &= \frac{-6}{(x+1)^2} \ \end{aligned}$

Step 4: $f'(x) = \frac{-6}{(x+1)^2}$ #### Rational + Radical (d) $f(x) = \frac{1}{\sqrt{x} - 4}$

Step 1: $f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

Step 2: $f(x+h) = \frac{1}{\sqrt{x+h} - 4}$

Step 3: $\begin{aligned} f'(x) &= \lim{h \to 0} \frac{\frac{1}{\sqrt{x+h} - 4} - \frac{1}{\sqrt{x} - 4}}{h} \ &= \lim{h \to 0} \frac{\frac{(\sqrt{x} - 4) - (\sqrt{x+h} - 4)}{(\sqrt{x+h} - 4)(\sqrt{x} - 4)}}{h} \ &= \lim{h \to 0} \frac{\sqrt{x} - \sqrt{x+h}}{h(\sqrt{x+h} - 4)(\sqrt{x} - 4)} \ &= \lim{h \to 0} \frac{(\sqrt{x} - \sqrt{x+h})(\sqrt{x} + \sqrt{x+h})}{h(\sqrt{x+h} - 4)(\sqrt{x} - 4)(\sqrt{x} + \sqrt{x+h})} \ &= \lim{h \to 0} \frac{x - (x+h)}{h(\sqrt{x+h} - 4)(\sqrt{x} - 4)(\sqrt{x} + \sqrt{x+h})} \ &= \lim{h \to 0} \frac{-h}{h(\sqrt{x+h} - 4)(\sqrt{x} - 4)(\sqrt{x} + \sqrt{x+h})} \ &= \lim_{h \to 0} \frac{-1}{(\sqrt{x+h} - 4)(\sqrt{x} - 4)(\sqrt{x} + \sqrt{x+h})} \ &= \frac{-1}{(\sqrt{x} - 4)(\sqrt{x} - 4)(\sqrt{x} + \sqrt{x})} \ &= \frac{-1}{2\sqrt{x}(\sqrt{x} - 4)^2} \ \end{aligned}$

Step 4: $f'(x) = \frac{-1}{2\sqrt{x}(\sqrt{x} - 4)^2}$ ### Rules of Derivative / Differentiation For $y = f(x)$ , the notation of the derivative is: $y' = f'(x) = \frac{dy}{dx} = \frac{d}{dx}f(x)$ #### Constant Rule Let $y = f(x) = a$ , where $a$ is a constant, then $f'(x) = \frac{dy}{dx} = 0$ #### Power Rule Let $y = f(x) = ax^n$ , then $\frac{dy}{dx} = f'(x) = nax^{n-1}$

Example 2.2 Differentiate the following with respect to x. a) $y = 10$ $\frac{dy}{dx} = 0$ b) $y = -x^5 + x^8 - 12$ $\frac{dy}{dx} = -5x^4 + 8x^7$ c) $y = \sqrt{x} = x^{\frac{1}{2}}$ $\frac{dy}{dx} = \frac{1}{2}x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}$ d) $y = -\frac{2}{3\sqrt{x}} = -\frac{2}{3}x^{-\frac{1}{2}}$ $\frac{dy}{dx} = -\frac{2}{3} \cdot \left(-\frac{1}{2}\right) x^{-\frac{3}{2}} = \frac{1}{3x^{\frac{3}{2}}}$ #### Product Rule If $y = f(x) \cdot g(x)$ , then let $u = f(x)$ and $v = g(x)$ $\frac{dy}{dx} = u'v + uv'$

Example 2.3 Differentiate the following with respect to x. (a) $y = (3x + 11)(2x - 7)$ Let $u = 3x + 11$ , $v = 2x - 7$ $u' = 3$ , $v' = 2$ $\frac{dy}{dx} = (3)(2x - 7) + (3x + 11)(2) = 6x - 21 + 6x + 22 = 12x + 1$ Alternative: Expand the $y$ . $y = 6x^2 - 21x + 22x - 77 = 6x^2 + x - 77$ $\frac{dy}{dx} = 12x + 1$ (b) $f(x) = (x^2 + 1)(x - 1)$ Let $u = x^2 + 1$ , $v = x - 1$ $u' = 2x$ , $v' = 1$ $f'(x) = (2x)(x - 1) + (x^2 + 1)(1) = 2x^2 - 2x + x^2 + 1 = 3x^2 - 2x + 1$ Alternative: Expand the $f(x)$ . $f(x) = x^3 - x^2 + x - 1$ $f'(x) = 3x^2 - 2x + 1$ #### Quotient Rule If $y = \frac{f(x)}{g(x)}$ , then let $u = f(x)$ and $v = g(x)$ $\frac{dy}{dx} = \frac{vu' - uv'}{v^2}$

Example 2.4 Differentiate the following with respect to x. (a) $y = \frac{2x + 3}{3x - 5}$ Let $u = 2x + 3$ , $v = 3x - 5$ $u' = 2$ , $v' = 3$ $\frac{dy}{dx} = \frac{(3x - 5)(2) - (2x + 3)(3)}{(3x - 5)^2} = \frac{6x - 10 - 6x - 9}{(3x - 5)^2} = \frac{-19}{(3x - 5)^2}$ (b) $f(x) = \frac{5x^2 + 11}{x - 4}$ Let $u = 5x^2 + 11$ , $v = x - 4$ $u' = 10x$ , $v' = 1$ $\frac{dy}{dx} = \frac{(x - 4)(10x) - (5x^2 + 11)(1)}{(x - 4)^2} = \frac{10x^2 - 40x - 5x^2 - 11}{(x - 4)^2} = \frac{5x^2 - 40x - 11}{(x - 4)^2}$ #### Derivative of Trigonometric Functions - $y = \sin(x)$ , $\frac{dy}{dx} = \cos(x)$ - $y = \cos(x)$ , $\frac{dy}{dx} = -\sin(x)$ - $y = \tan(x)$ , $\frac{dy}{dx} = \sec^2(x)$ - $y = \sec(x)$ , $\frac{dy}{dx} = \sec(x)\tan(x)$ - $y = \csc(x)$ , $\frac{dy}{dx} = -\csc(x)\cot(x)$ - $y = \cot(x)$ , $\frac{dy}{dx} = -\csc^2(x)$

Example 2.5 Find the derivative of the following functions. a) $y = 2\ln(x) - \sin(x) + x e^3$ $\frac{dy}{dx} = \frac{2}{x} - \cos(x) + e^3$ b) $y = \frac{1}{x} - 3\sec(x) - \ln(6) + 5x$ $\frac{dy}{dx} = -\frac{1}{x^2} - 3\sec(x)\tan(x) + 5$ c) $y = \ln(x)\tan(x)$ Let $u = \ln(x)$ , $v = \tan(x)$ $u' = \frac{1}{x}$ , $v' = \sec^2(x)$ $\frac{dy}{dx} = \frac{1}{x}\tan(x) + \ln(x)\sec^2(x)$ d) $y = \frac{e^x}{\cos(x)}$ Let $u = e^x$ , $v = \cos(x)$ $u' = e^x$ , $v' = -\sin(x)$ $\frac{dy}{dx} = \frac{\cos(x)e^x - e^x(-\sin(x))}{\cos^2(x)} = \frac{e^x(\cos(x) + \sin(x))}{\cos^2(x)}$ #### Chain Rule If $g$ is differentiable at $x$ and $f$ is differentiable at $g(x)$ , then the composite function $F(x) = f(g(x))$ has the derivative: $F'(x) = f'(g(x)) \cdot g'(x)$ In Leibniz notation, if $y = f(u)$ and $u = g(x)$ , then $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Example 2.6 Find $F'(x)$ if a) $F(x) = (x^2 - 3)^2$ Composite Function $F(x) = f(g(x))$ Let $f(u) = u^2$ and $g(x) = x^2 - 3$ $f'(u) = 2u$ and $g'(x) = 2x$ $F'(x) = f'(g(x)) \cdot g'(x) = 2(x^2 - 3)(2x) = 4x(x^2 - 3)$ Alternative using Chain Rule: Let $y = u^2$ where $u = x^2 - 3$ $\frac{dy}{du} = 2u$ and $\frac{du}{dx} = 2x$ $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = (2u)(2x) = 2(x^2 - 3)(2x) = 4x(x^2 - 3)$ b) $F(x) = \sqrt[9]{x - 1} = (x - 1)^{\frac{1}{9}}$ Composite Function $F(x) = f(g(x))$ Let $f(u) = u^{\frac{1}{9}}$ and $g(x) = x - 1$ $f'(u) = \frac{1}{9}u^{-\frac{8}{9}}$ and $g'(x) = 1$ $F'(x) = f'(g(x)) \cdot g'(x) = \frac{1}{9}(x - 1)^{-\frac{8}{9}}(1) = \frac{1}{9(x - 1)^{\frac{8}{9}}}$ Alternative using Chain Rule: Let $y = u^{\frac{1}{9}}$ where $u = x - 1$ $\frac{dy}{du} = \frac{1}{9}u^{-\frac{8}{9}}$ and $\frac{du}{dx} = 1$ $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{9}u^{-\frac{8}{9}}(1) = \frac{1}{9(x - 1)^{\frac{8}{9}}}$ Note: By using the chain rule method, the working step is from the outside to the inside. $\frac{d}{dx}f(g(x)) = f'(g(x)) \cdot g'(x)$ With the combination of the Power Rule and the Chain Rule, the derivative process is as below: If $y = [u]^n$ , then $\frac{dy}{dx} = n[u]^{n-1} \cdot f'(u)$ If $y = [f(x)]^n$ , then $\frac{dy}{dx} = n[f(x)]^{n-1} \cdot f'(x)$ If $y = [f(g(x))]^n$ , then $\frac{dy}{dx} = n[f(g(x))]^{n-1} \cdot f'(g(x)) \cdot g'(x)$ #### Polynomial, Rational, Radical Functions

Example 2.7 Differentiate the following with respect to x. a) $y = \sqrt{4x^3}$ $\frac{dy}{dx} = \frac{3}{2}(4x^3)^{-\frac{1}{2}} \cdot (12x^2) = \frac{18x^2}{\sqrt{4x^3}}$ b) $y = \frac{5}{\sqrt[9]{x^4}} = 5x^{-\frac{4}{9}}$ $\frac{dy}{dx} = -\frac{20}{9}x^{-\frac{13}{9}} = -\frac{20}{9\sqrt[9]{x^{13}}}$ c) $y = (x^2 + 10)^{3}$ $\frac{dy}{dx} = 3(x^2 + 10)^{2} \cdot (2x) = 6x(x^2 + 10)^2$ d) $y = (x + 1)^3 (x - 1)$ Product Rule: $\frac{dy}{dx} = u'v + uv'$ Let $u = (x+1)^3$ , $v = x-1$ $u' = 3(x+1)^2$ , $v' = 1$ $\frac{dy}{dx} = 3(x+1)^2 (x-1) + (x+1)^3 (1) = (x+1)^2[3(x-1) + (x+1)] = (x+1)^2[3x - 3 + x + 1] = (x+1)^2(4x - 2) = 2(x+1)^2(2x - 1)$ #### Trigonometry Functions

Example 2.8 Differentiate the following with respect to x. a) $y = \sin(5x)$ $\frac{dy}{dx} = \cos(5x) \cdot 5 = 5\cos(5x)$ b) $y = \cos(\sin(x))$ $\frac{dy}{dx} = -\sin(\sin(x)) \cdot \cos(x) = -\cos(x)\sin(\sin(x))$ c) $y = \tan^6(3x)$ $\frac{dy}{dx} = 6\tan^5(3x) \cdot \sec^2(3x) \cdot 3 = 18\sec^2(3x)\tan^5(3x)$ d) $f(x) = \tan \left[\frac{2 - x^2}{2 + x^2}\right]$ $$\begin{aligned} f'(x) &= \sec^2 \left[\frac{2 - x^2}{2 + x^2}\right] \cdot \left[\frac{(2 + x^2)(-2x) - (2 - x^2)(2x)}{(2 + x^2)^2}\right] \ &= \sec^2 \left[\frac{2 - x^2}{2 + x^2}\right] \cdot \left[\frac{-