Differentiation

Starter

Q

$$\begin{align} 2x + \sqrt{ x } - 6 &= 0 \\ \text{Let }y = \sqrt{ x } \\ 2y^2 + y - 6 &= 0 \\ y &= \frac{-1 \pm \sqrt{ 1^2 - 4 \cdot 2 \cdot -6 }}{ 2 \cdot 2 } \\ y &= 1.5 \lor \underbrace{ \cancel{ -2 } }_{ \text{Cannot be negative} } \\ \\ y &= \sqrt{ x } \\[4pt] 1.5 &= \sqrt{ x } \\ 2.25 &= x \\[4pt] \\ \therefore x = 2.25 \lor 4 \end{align}$$ $$\begin{align} x^{8} - 17x^{4} + 16 &= 0 \\ \text{Let }y = x^4 \\ y^2 - 17x + 16 &= 0 \\ 0 &= (y-16)(y-1) \\ y &= 16 \lor 1 \\ \\ x^{4} &= 16 \\ x &= \pm \sqrt[4]{ 16 } \\ x &= \pm 2 \\[4pt] x^{4} &= 1 \\ x &= \pm 1 \\ \\ \therefore x = \pm 2, \pm 1 \end{align}$$

12 (EP)

$f(x) = 4kx^2 + (4k+2)x + 1$, where $k$ is a real constant. Find the discriminant in terms of k

$$\begin{align} 16k^2 + 4 \end{align}$$

Functions

y=x^2+1

$$\begin{gather} \text{Domain} = \mathbb{R} \\ \text{Range} = \{y \in \mathbb{R} | y \geq 1 \} \end{gather}$$

Example 8

The functions $f$ and $g$ are given by $f(x) = 2x - 10$ and $g(x) = x^2 - 9,\ x \in \mathbb{R}$.

Find the values of $f(5)$ and $g(10)$

$$\begin{align} f(5) &= 2 \cdot 5 - 10 \\ &= 0 \\ \\ g(10) &= 10^2 - 9 \\ &= 100 - 9 \\ &= 91 \end{align}$$

Example 9

$f(x) = x^2 + 6x - 5,\ x \in \mathbb{R}$

$$\begin{align} x^2 + 6x - 5 &= (x + 3)^2 - 9 - 5 \\ &= (x + 3)^2 - 14 \\ \\ \text{Roots} \\ 0 &= (x+3)^2 - 14 \\ 14 &= (x+3)^2 \\ \pm \sqrt{ 14 } &= x+3 \\ -3 \pm \sqrt{ 14 } &= x \\ \\ \text{Minimum value} \\ (x + 3)^2 &\geq 0 \\ (x + 3)^2 - 14 &\geq -14 \\ \therefore \text{Minimum is } -14 \end{align}$$

Gradients of curves

left=-10; right=10;
bottom=-5; top=30;
---
x^2|black
(5, 25)|blue|label:B
(4.5, 20.25)|blue|label:B
(4.1, 16.81)|blue|label:B
(4.01, 16.0801)|blue|label:B
(4,16)|black|cross|label:A

Find gradients of the chords joining the point $(4, 16)$ to the points with coordinates:

$$\begin{gathered} \begin{align} (5, 25) \\ m &= \frac{25-16}{5-4} \\ &= 9 \\ \\ (4.5, 20.25) \\ m &= \frac{20.25-16}{4.5-4} \\ &= 8.5 \\ \\ (4.1, 16.81) \\ m &= \frac{16.81-16}{4.1-4} \\ &= 8.10000000000002 \\ \\ (4.01, 16.0801) \\ m &= \frac{16.0801-16}{4.01-4} \\ &= 8.01000000000033 \\ \\ (4+h, (4+h)^2) \\ m &= \frac{(4+h)^2 - 16}{4 + h - 4} \\ &= \frac{h^2 + 8h + 16 - 16}{h} \\ &= \frac{h^2 + 8h}{h} \\ &= h + 8\\ \end{align} \\[4pt] \lim_{ h \to 0 } (h + 8)=8 \\ \therefore \text{Gradient is }8 \end{gathered}$$

Example 2

In this example, we work out the gradient at a specific point. This means we aim for a specific numerical value (gradient at a point).

The point $A$ with coordinates $(4, 16)$ lies on the curve with equation $y=x^2$. At point $A$ the curve has the gradient $g$.

Show that $g = \lim_{ h \to 0 }(8+h)$.

$$\begin{align} g &= \lim_{ h \to 0 } \frac{f(4+h)-f(4)}{h} \\ &= \lim_{ h \to 0 } \frac{(4+h)^2 - 4^2}{h} \\ &= \lim_{ h \to 0 } \frac{16 + 8h + h^2 - 16}{h} \\ &= \lim_{ h \to 0 } \frac{8h + h^2}{h} \\ &= \lim_{ h \to 0 } (8+h) \end{align}$$

Deduce the value of $g$.

$$g=8$$

Example 3

In this example, we work out the gradient at any random point of the curve. That means we aim for an expression in terms of $x$ (gradient function). This can be used to evaluate (by substituting the point’s $x$ coordinate) the gradient at any point on the graph.

Prove, from first principles, that the derivative of $x^3$ is $3x^2$

$$\begin{gathered} \begin{align} A &= (x, x^3) \\ B &= (x+h, (x+h)^3) \\ \\ m = f'(x) &= \lim_{ h \to 0 } \frac{(x+h)^3 - x^3}{x+h - x} \\ &= \lim_{ h \to 0 } \frac{x^3 + 3x^2h + 3xh^2 + h^3 + x^3}{h} \\ &= \lim_{ h \to 0 } \frac{3x^2h + 3xh^2 + h^3}{h} \\ &= \lim_{ h \to 0 } \frac{h(3x^2 + 3xh + h^2)}{h} \\ &= \lim_{ h \to 0 } ({3x^2 + 3xh + h^2}) \end{align} \\[4pt] \text{As }h \to 0, 3xh \to 0\text{ and }h^2 \to 0 \\ \text{So $f'(x) = 3x^2$} \end{gathered}$$

General Derivitive Formula

$$(x^{n})^{1} = nx^{n-1}$$

Because:

$$\lim_{ h \to 0 } \frac{(x+h)^{n} - x^{n}}{h} = nx^{n-1}$$

For all real values of $n$, and for constant $a$:

• If $f(x)=x^{n}$ then $f'(x) = nx^{n-1}$ If $y = x^{n}$ then $\frac{dy}{dx} = nx^{n-1}$
• If $f(x) = ax^{n}$ then $f'(x) = anx^{n-1}$ If $y=ax^{n}$ then $\frac{dy}{dx} = anx^{n-1}$

Challenge

Differentiating Quadratics

If $y = f(x) \pm g(x)$, then $\frac{dy}{dx} = f'(x) \pm g'(x)$

Gradients, Tangents, and Normals

The equation of a straight line with gradient $m$ that passes through the point $(x_{1}, y_{1})$ is $y-y_{1} = m(x-x_{1})$

You can use the derivative to find the equation of the tangent to a curve at a given point. On the curve with the equation $y=f(x)$, the gradient of the tangent at a point $A$ with $x$-coordinate $a$ will be $f'(a)$

The gradient of a curve at $A(x_{1}, y_{1})$:

$$\begin{gather} m=\frac{y-y_{1}}{x-x_{1}} \\ f=f'(x_{2}) \\ m=\tan \theta \end{gather}$$

The tangent to the curve $y=f(x)$ at the point with coordinates $(a, f(a))$ has equation

$$y-f(a)=f'(a)(x-a)$$

The normal to the curve $y=f(x)$ at the point with coordinates $(a, f(a))$ has equation

$$y-f(a)=-\frac{1}{f'(a)}(x-a)$$

Increasing and Decreasing Functions

The function $f(x)$ is increasing on the interval $[a, b]$ if $f'(x) \geq 0$ for all values of $x$ such that a < x < b. The function $f(x)$ is decreasing on the interval $[a, b]$ if $f'x(x) \leq 0$ for all values of $x$ such that a < x < b

y=x^4 - 2x^2
(-1, -1)
(1, -1)

The function $f(x) = x^{4} - 2x^2$ is increasing where $[-1, 0] \cup [1, +\infty)$, and