# Proof Derivative of Quadratic Functions

The proof of the derivative of the quadratic functions is presented using the limit definition of the derivative .

## Proof of the Derivative the Quadratic Function Using Definition of the Derivative

The definition of the derivative $f'$ of a function $f$ is given by the limit $f'(x) = \lim_{h \to 0} \dfrac{f(x+h)-f(x)}{h}$
Let $f$ be a quadratic function of the form: $f(x) = a x^2 + bx + c$ and write the derivative of $f$ as follows
$f'(x) = \lim_{h \to 0} \dfrac{ a (x+h)^2 + b(x+h) + c - (a x^2 + bx + c)}{h}$

Expand the terms $a (x+h)^2$ and $b(x+h)$ in the numerator
$f'(x) = \lim_{h \to 0} \dfrac{ a x^2 + a h^2 + 2 a x h + b x + bh + c - a x^2 - bx - c)}{h}$

Simplify the numerator
$f'(x) = \lim_{h \to 0} \dfrac{ a h^2 + 2 a x h + bh )}{h}$

Divide numerator and denomianator by $h$
$f'(x) = \lim_{h \to 0} a h + 2 a x + b$

Evaluate the limit to obtain the derivative of the quadratic function as
$f'(x) = 2 a x + b$

## Examples with Solutions

Part A
Find the derivatives of the quadratic functions given by
a) $f(x) = 4x^2 - x + 1$
b) $g(x) = - x^2 - 1$
c) $h(x) = 0.1 x^2 - \dfrac {x}{2} - 100$
d) $f(x) = - \dfrac { 3 x^2}{7} - 0.2 x + 7$

Part B
Let $f(x) = a x^2 + b x + c$. Find $f'(2)$ given that $f(2) = 3$, $f'(0) = 1$ and $f'(-1) = 2$

## Solutions to the Above Examples

Part A
a) $f'(x) = 8 x - 1$
b) $g'(x) = - 2 x$
c) $h'(x) = 0.2 x - \dfrac {1}{2}$
d) $f'(x) = -\dfrac {6 x}{7} - 0.2$

Part B
The given function $f$ is a quadratic function, hence
$f'(x) = 2 a x + b$
Given $f'(0) = 1$, substitute to obtain the equation: $2 a (0) + b = 1$
Solve for $b$ to obtain: $b = 1$
Given $f'(-1) = 2$, substitute to obtain the equation: $2 a (-1) + 1 = 2$
Solve for $a$ to obtain: $a = -\dfrac{1}{2}$
Given $f(2) = 3$, substitute to obtain the equation: $f(2) = -\dfrac{1}{2} (2)^2 + (1) (2) + c = 3$
Solve for $c$ to obatain: $c = 3$
$f'(2) = 2 (-\dfrac{1}{2})(2) + 1 = -1$

2. Proof of Derivative of $\ln(x)$.
3. Proof of Derivative of $cos(x)$.
4. Proof of Derivative of $sin(x)$.