Rate of change
Rate of Change Using Derivatives:
 Derivatives in calculus measure the rate at which one quantity changes concerning another, representing instantaneous rates of change.
Derivative as Rate of Change:
 Function Derivative: The derivative ${f}^{\mathrm{\prime}}(x)$ of a function $f(x)$ at a point represents the rate of change of $f(x)$ concerning $x$ at that point.
 Average Rate of Change: $\frac{\mathrm{\Delta}y}{\mathrm{\Delta}x}$ represents the average rate of change of $y$ concerning $x$ over an interval.
 Limit as Delta $\to $ 0: The derivative represents the instantaneous rate of change as the interval approaches zero.
Example Problem:
Problem Statement:
Consider a function $f(x)=3{x}^{2}+2x5$. Find the rate of change of $f(x)$ concerning $x$ at $x=2$.
Solution Steps:

Find the Derivative ${f}^{\mathrm{\prime}}(x)$:
 $f(x)=3{x}^{2}+2x5$
 ${f}^{\mathrm{\prime}}(x)=\frac{d}{dx}[3{x}^{2}+2x5]$
 ${f}^{\mathrm{\prime}}(x)=6x+2$

Evaluate the Derivative at $x=2$:
 ${f}^{\mathrm{\prime}}(x)=6x+2$
 ${f}^{\mathrm{\prime}}(2)=6(2)+2=12+2=14$

Interpretation:
 The rate of change of $f(x)$ concerning $x$ at $x=2$is $14$.
Interpretation:
 Positive Rate of Change: Indicates an increasing function concerning $x$ at that point.
 Negative Rate of Change: Indicates a decreasing function concerning $x$ at that point.
Example:
Problem Statement: Suppose the position of an object moving along a straight line is given by the function $s(t)=3{t}^{2}2t+5$, where $s$ represents position in meters and $t$ represents time in seconds. Find the object's velocity and acceleration functions.
Solution Steps:

Velocity Function:
 The velocity of the object is the rate of change of position concerning time.
 Calculate the derivative of the position function $s(t)$ to find the velocity function $v(t)$.
 $v(t)={s}^{\mathrm{\prime}}(t)=\frac{ds}{dt}=\frac{d}{dt}(3{t}^{2}2t+5)$
 $v(t)=6t2$ meters per second (m/s)

Acceleration Function:
 The acceleration of the object is the rate of change of velocity concerning time.
 Calculate the derivative of the velocity function $v(t)$ to find the acceleration function $a(t)$.
 $a(t)={v}^{\mathrm{\prime}}(t)=\frac{dv}{dt}=\frac{d}{dt}(6t2)$
 $a(t)=6$ meters per second squared (m/s²)
Interpretation:
 Velocity: The object's velocity is given by $v(t)=6t2$. It changes at a rate of $6$m/s concerning time.
 Acceleration: The object's acceleration is a constant $6$m/s². It represents the rate at which the velocity changes concerning time.
Example:
Problem Statement: Consider the function $f(x)={x}^{2}+3x5$. Find the average rate of change of $f(x)$ concerning $x$ between $x=2$ and $x=4$.
Solution Steps:

Calculate $f(x)$ at $x=2$ and $x=4$:
 $f(2)={2}^{2}+3(2)5=4+65=5$
 $f(4)={4}^{2}+3(4)5=16+125=23$

Calculate Average Rate of Change:
 Average Rate of Change $=\frac{\mathrm{\Delta}y}{\mathrm{\Delta}x}=\frac{f(4)f(2)}{42}$
 Average Rate of Change $=\frac{235}{42}=\frac{18}{2}=9$
Interpretation:
 The average rate of change of $f(x)={x}^{2}+3x5$ concerning $x$ between $x=2$ and $x=4$ is $9$ units.
Applications:
 Physics: Analyzing velocity, acceleration, and force concerning time.
 Economics: Determining marginal cost, revenue, and profit concerning production levels or sales.
 Engineering: Optimizing designs, analyzing structural stability, and describing rates of flow.
Key Points:
 Tangent Line: The slope of the tangent line to the graph of a function at a point represents the instantaneous rate of change at that point.
 Derivative Calculation: Derivatives are calculated using differentiation rules to find the rate of change of a function concerning its variable.