Differentiation: Gradients, tangents, normals, rate of change.
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Lesson Objectives
- Understand differentiation as the process of finding the derivative
- Differentiate from first principles using the limit definition
- Apply the power rule, sum/difference rule, and constant multiple rule
- Find the gradient of a curve at a given point
- Determine the equation of a tangent line to a curve
- Determine the equation of a normal line to a curve
- Interpret the derivative as the instantaneous rate of change
Introduction to Differentiation
Differentiation is a branch of calculus that deals with rates of change. The derivative of a function measures how the function changes as its input changes. It represents the slope (gradient) of the tangent line to the curve at any point. Differentiation has applications in physics, engineering, economics, biology, and many other fields.
The Derivative
The derivative of $f(x)$ with respect to $x$ is denoted as $f'(x)$ or $\frac{dy}{dx}$.
It represents the instantaneous rate of change or the gradient of the tangent line at a point.
The derivative of $f(x)$ with respect to $x$ is denoted as $f'(x)$ or $\frac{dy}{dx}$.
It represents the instantaneous rate of change or the gradient of the tangent line at a point.
Key Definitions:
• Derivative: The rate at which a function changes at a given point.
• Differentiation: The process of finding the derivative.
• Gradient of a Curve: The slope of the tangent line at a point on the curve.
• Tangent Line: A line that touches the curve at exactly one point.
• Normal Line: A line perpendicular to the tangent line at the point of contact.
• Rate of Change: How quickly one quantity changes relative to another.
• Derivative: The rate at which a function changes at a given point.
• Differentiation: The process of finding the derivative.
• Gradient of a Curve: The slope of the tangent line at a point on the curve.
• Tangent Line: A line that touches the curve at exactly one point.
• Normal Line: A line perpendicular to the tangent line at the point of contact.
• Rate of Change: How quickly one quantity changes relative to another.
Differentiation from First Principles
The derivative of a function $f(x)$ is defined as the limit of the difference quotient as $\Delta x$ approaches zero: $$f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$$ This is called differentiation from first principles, where $\Delta x$ represents a small change in $x$.
Step-by-Step Method for First Principles:
1. Write the definition: $f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$
2. Substitute $f(x + \Delta x)$ and $f(x)$.
3. Simplify the numerator by expanding and cancelling terms.
4. Factor out $\Delta x$ from the numerator.
5. Cancel $\Delta x$ (since $\Delta x \neq 0$ in the limit).
6. Take the limit as $\Delta x \to 0$ to find $f'(x)$.
1. Write the definition: $f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$
2. Substitute $f(x + \Delta x)$ and $f(x)$.
3. Simplify the numerator by expanding and cancelling terms.
4. Factor out $\Delta x$ from the numerator.
5. Cancel $\Delta x$ (since $\Delta x \neq 0$ in the limit).
6. Take the limit as $\Delta x \to 0$ to find $f'(x)$.
Example 1: Differentiating f(x) = x²
Problem: Find $f'(x)$ from first principles if $f(x) = x^2$.
Formula: $f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$
Solution:
Step 1: Write the definition: $f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$
Step 2: Substitute $f(x+\Delta x) = (x+\Delta x)^2$ and $f(x) = x^2$:
$f'(x) = \lim_{\Delta x \to 0} \frac{(x+\Delta x)^2 - x^2}{\Delta x}$
Step 3: Expand $(x+\Delta x)^2 = x^2 + 2x\Delta x + (\Delta x)^2$:
$f'(x) = \lim_{\Delta x \to 0} \frac{x^2 + 2x\Delta x + (\Delta x)^2 - x^2}{\Delta x}$
Step 4: Simplify the numerator: $x^2 - x^2 = 0$, leaving:
$f'(x) = \lim_{\Delta x \to 0} \frac{2x\Delta x + (\Delta x)^2}{\Delta x}$
Step 5: Factor out $\Delta x$ from the numerator:
$f'(x) = \lim_{\Delta x \to 0} \frac{\Delta x(2x + \Delta x)}{\Delta x}$
Step 6: Cancel $\Delta x$ (since $\Delta x \neq 0$ in the limit):
$f'(x) = \lim_{\Delta x \to 0} (2x + \Delta x)$
Step 7: Take the limit as $\Delta x \to 0$:
$f'(x) = 2x + 0 = 2x$
Answer: $f'(x) = 2x$
Problem: Find $f'(x)$ from first principles if $f(x) = x^2$.
Formula: $f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$
Solution:
Step 1: Write the definition: $f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$
Step 2: Substitute $f(x+\Delta x) = (x+\Delta x)^2$ and $f(x) = x^2$:
$f'(x) = \lim_{\Delta x \to 0} \frac{(x+\Delta x)^2 - x^2}{\Delta x}$
Step 3: Expand $(x+\Delta x)^2 = x^2 + 2x\Delta x + (\Delta x)^2$:
$f'(x) = \lim_{\Delta x \to 0} \frac{x^2 + 2x\Delta x + (\Delta x)^2 - x^2}{\Delta x}$
Step 4: Simplify the numerator: $x^2 - x^2 = 0$, leaving:
$f'(x) = \lim_{\Delta x \to 0} \frac{2x\Delta x + (\Delta x)^2}{\Delta x}$
Step 5: Factor out $\Delta x$ from the numerator:
$f'(x) = \lim_{\Delta x \to 0} \frac{\Delta x(2x + \Delta x)}{\Delta x}$
Step 6: Cancel $\Delta x$ (since $\Delta x \neq 0$ in the limit):
$f'(x) = \lim_{\Delta x \to 0} (2x + \Delta x)$
Step 7: Take the limit as $\Delta x \to 0$:
$f'(x) = 2x + 0 = 2x$
Answer: $f'(x) = 2x$
Example 2: Differentiating f(x) = 3x²
Problem: Find $f'(x)$ from first principles if $f(x) = 3x^2$.
Formula: $f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$
Solution:
Step 1: Write the definition: $f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$
Step 2: Substitute $f(x+\Delta x) = 3(x+\Delta x)^2$ and $f(x) = 3x^2$:
$f'(x) = \lim_{\Delta x \to 0} \frac{3(x+\Delta x)^2 - 3x^2}{\Delta x}$
Step 3: Expand $(x+\Delta x)^2 = x^2 + 2x\Delta x + (\Delta x)^2$:
$f'(x) = \lim_{\Delta x \to 0} \frac{3(x^2 + 2x\Delta x + (\Delta x)^2) - 3x^2}{\Delta x}$
Step 4: Distribute the 3:
$f'(x) = \lim_{\Delta x \to 0} \frac{3x^2 + 6x\Delta x + 3(\Delta x)^2 - 3x^2}{\Delta x}$
Step 5: Simplify the numerator: $3x^2 - 3x^2 = 0$, leaving:
$f'(x) = \lim_{\Delta x \to 0} \frac{6x\Delta x + 3(\Delta x)^2}{\Delta x}$
Step 6: Factor out $\Delta x$ from the numerator:
$f'(x) = \lim_{\Delta x \to 0} \frac{\Delta x(6x + 3\Delta x)}{\Delta x}$
Step 7: Cancel $\Delta x$ (since $\Delta x \neq 0$ in the limit):
$f'(x) = \lim_{\Delta x \to 0} (6x + 3\Delta x)$
Step 8: Take the limit as $\Delta x \to 0$:
$f'(x) = 6x + 0 = 6x$
Answer: $f'(x) = 6x$
Problem: Find $f'(x)$ from first principles if $f(x) = 3x^2$.
Formula: $f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$
Solution:
Step 1: Write the definition: $f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$
Step 2: Substitute $f(x+\Delta x) = 3(x+\Delta x)^2$ and $f(x) = 3x^2$:
$f'(x) = \lim_{\Delta x \to 0} \frac{3(x+\Delta x)^2 - 3x^2}{\Delta x}$
Step 3: Expand $(x+\Delta x)^2 = x^2 + 2x\Delta x + (\Delta x)^2$:
$f'(x) = \lim_{\Delta x \to 0} \frac{3(x^2 + 2x\Delta x + (\Delta x)^2) - 3x^2}{\Delta x}$
Step 4: Distribute the 3:
$f'(x) = \lim_{\Delta x \to 0} \frac{3x^2 + 6x\Delta x + 3(\Delta x)^2 - 3x^2}{\Delta x}$
Step 5: Simplify the numerator: $3x^2 - 3x^2 = 0$, leaving:
$f'(x) = \lim_{\Delta x \to 0} \frac{6x\Delta x + 3(\Delta x)^2}{\Delta x}$
Step 6: Factor out $\Delta x$ from the numerator:
$f'(x) = \lim_{\Delta x \to 0} \frac{\Delta x(6x + 3\Delta x)}{\Delta x}$
Step 7: Cancel $\Delta x$ (since $\Delta x \neq 0$ in the limit):
$f'(x) = \lim_{\Delta x \to 0} (6x + 3\Delta x)$
Step 8: Take the limit as $\Delta x \to 0$:
$f'(x) = 6x + 0 = 6x$
Answer: $f'(x) = 6x$
Example 3: Differentiating a Linear Function
Problem: Find $f'(x)$ from first principles if $f(x) = 3x + 2$.
Formula: $f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$
Solution:
$f'(x) = \lim_{\Delta x \to 0} \frac{[3(x+\Delta x)+2] - [3x+2]}{\Delta x} = \lim_{\Delta x \to 0} \frac{3x+3\Delta x+2 - 3x - 2}{\Delta x} = \lim_{\Delta x \to 0} \frac{3\Delta x}{\Delta x} = \lim_{\Delta x \to 0} 3 = 3$
Answer: $f'(x) = 3$
Problem: Find $f'(x)$ from first principles if $f(x) = 3x + 2$.
Formula: $f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$
Solution:
$f'(x) = \lim_{\Delta x \to 0} \frac{[3(x+\Delta x)+2] - [3x+2]}{\Delta x} = \lim_{\Delta x \to 0} \frac{3x+3\Delta x+2 - 3x - 2}{\Delta x} = \lim_{\Delta x \to 0} \frac{3\Delta x}{\Delta x} = \lim_{\Delta x \to 0} 3 = 3$
Answer: $f'(x) = 3$
Example 4: Differentiating a Cubic Function
Problem: Find $f'(x)$ from first principles if $f(x) = x^3$.
Formula: $f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$
Solution:
$f'(x) = \lim_{\Delta x \to 0} \frac{(x+\Delta x)^3 - x^3}{\Delta x} = \lim_{\Delta x \to 0} \frac{x^3 + 3x^2\Delta x + 3x(\Delta x)^2 + (\Delta x)^3 - x^3}{\Delta x}$
$= \lim_{\Delta x \to 0} \frac{3x^2\Delta x + 3x(\Delta x)^2 + (\Delta x)^3}{\Delta x} = \lim_{\Delta x \to 0} (3x^2 + 3x\Delta x + (\Delta x)^2) = 3x^2$
Answer: $f'(x) = 3x^2$
Problem: Find $f'(x)$ from first principles if $f(x) = x^3$.
Formula: $f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$
Solution:
$f'(x) = \lim_{\Delta x \to 0} \frac{(x+\Delta x)^3 - x^3}{\Delta x} = \lim_{\Delta x \to 0} \frac{x^3 + 3x^2\Delta x + 3x(\Delta x)^2 + (\Delta x)^3 - x^3}{\Delta x}$
$= \lim_{\Delta x \to 0} \frac{3x^2\Delta x + 3x(\Delta x)^2 + (\Delta x)^3}{\Delta x} = \lim_{\Delta x \to 0} (3x^2 + 3x\Delta x + (\Delta x)^2) = 3x^2$
Answer: $f'(x) = 3x^2$
Watch Out!
When using first principles, be careful with algebraic manipulation. Expand $(x+\Delta x)^n$ correctly using the binomial theorem. The limit process removes terms with $\Delta x$.
When using first principles, be careful with algebraic manipulation. Expand $(x+\Delta x)^n$ correctly using the binomial theorem. The limit process removes terms with $\Delta x$.
Practice for First Principles
- Find $f'(x)$ from first principles if $f(x) = 5x - 3$.
- Find $f'(x)$ from first principles if $f(x) = x^2 + 2x$.
- Find $f'(x)$ from first principles if $f(x) = 4x^2$.
- Find $f'(x)$ from first principles if $f(x) = 2x^2 + 3x$.
- Find $f'(x)$ from first principles if $f(x) = 7$ (constant function).
Rules of Differentiation
Basic Differentiation Rules
1. Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$
2. Constant Multiple Rule: $\frac{d}{dx}(cf(x)) = c f'(x)$
3. Sum/Difference Rule: $\frac{d}{dx}(f(x) \pm g(x)) = f'(x) \pm g'(x)$
4. Product Rule: $\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)$
5. Quotient Rule: $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$
6. Chain Rule: $\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)$
1. Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$
2. Constant Multiple Rule: $\frac{d}{dx}(cf(x)) = c f'(x)$
3. Sum/Difference Rule: $\frac{d}{dx}(f(x) \pm g(x)) = f'(x) \pm g'(x)$
4. Product Rule: $\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)$
5. Quotient Rule: $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}$
6. Chain Rule: $\frac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)$
Example 1: Power Rule
Problem: Find $\frac{dy}{dx}$ if $y = x^5$.
Formula: $\frac{d}{dx}(x^n) = nx^{n-1}$
Solution:
$\frac{dy}{dx} = 5x^{5-1} = 5x^4$
Answer: $5x^4$
Problem: Find $\frac{dy}{dx}$ if $y = x^5$.
Formula: $\frac{d}{dx}(x^n) = nx^{n-1}$
Solution:
$\frac{dy}{dx} = 5x^{5-1} = 5x^4$
Answer: $5x^4$
Example 2: Sum and Difference Rules
Problem: Find $\frac{dy}{dx}$ if $y = 3x^4 - 2x^3 + 5x - 7$.
Formula: $\frac{d}{dx}(f(x) \pm g(x)) = f'(x) \pm g'(x)$
Solution:
$\frac{dy}{dx} = 3 \cdot 4x^3 - 2 \cdot 3x^2 + 5 \cdot 1 - 0 = 12x^3 - 6x^2 + 5$
Answer: $12x^3 - 6x^2 + 5$
Problem: Find $\frac{dy}{dx}$ if $y = 3x^4 - 2x^3 + 5x - 7$.
Formula: $\frac{d}{dx}(f(x) \pm g(x)) = f'(x) \pm g'(x)$
Solution:
$\frac{dy}{dx} = 3 \cdot 4x^3 - 2 \cdot 3x^2 + 5 \cdot 1 - 0 = 12x^3 - 6x^2 + 5$
Answer: $12x^3 - 6x^2 + 5$
Example 3: Product Rule
Problem: Find $\frac{dy}{dx}$ if $y = (2x + 1)(x^2 - 3)$.
Formula: $\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)$
Solution:
Let $f(x) = 2x + 1$, $f'(x) = 2$; $g(x) = x^2 - 3$, $g'(x) = 2x$
$y' = (2)(x^2 - 3) + (2x + 1)(2x) = 2x^2 - 6 + 4x^2 + 2x = 6x^2 + 2x - 6$
Answer: $6x^2 + 2x - 6$
Problem: Find $\frac{dy}{dx}$ if $y = (2x + 1)(x^2 - 3)$.
Formula: $\frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)$
Solution:
Let $f(x) = 2x + 1$, $f'(x) = 2$; $g(x) = x^2 - 3$, $g'(x) = 2x$
$y' = (2)(x^2 - 3) + (2x + 1)(2x) = 2x^2 - 6 + 4x^2 + 2x = 6x^2 + 2x - 6$
Answer: $6x^2 + 2x - 6$
Example 4: Quotient Rule
Problem: Find $f'(x)$ if $f(x) = \frac{x^2}{x+1}$.
Formula: $\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f'g - fg'}{g^2}$
Solution:
$f'(x) = \frac{(2x)(x+1) - (x^2)(1)}{(x+1)^2} = \frac{2x^2 + 2x - x^2}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2}$
Answer: $\frac{x^2 + 2x}{(x+1)^2}$
Problem: Find $f'(x)$ if $f(x) = \frac{x^2}{x+1}$.
Formula: $\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f'g - fg'}{g^2}$
Solution:
$f'(x) = \frac{(2x)(x+1) - (x^2)(1)}{(x+1)^2} = \frac{2x^2 + 2x - x^2}{(x+1)^2} = \frac{x^2 + 2x}{(x+1)^2}$
Answer: $\frac{x^2 + 2x}{(x+1)^2}$
Example 5: Chain Rule
Problem: Find $\frac{dy}{dx}$ if $y = (3x^2 + 1)^4$.
Formula: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
Solution:
Let $u = 3x^2 + 1$, then $y = u^4$
$\frac{dy}{du} = 4u^3$, $\frac{du}{dx} = 6x$
$\frac{dy}{dx} = 4(3x^2 + 1)^3 \cdot 6x = 24x(3x^2 + 1)^3$
Answer: $24x(3x^2 + 1)^3$
Problem: Find $\frac{dy}{dx}$ if $y = (3x^2 + 1)^4$.
Formula: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
Solution:
Let $u = 3x^2 + 1$, then $y = u^4$
$\frac{dy}{du} = 4u^3$, $\frac{du}{dx} = 6x$
$\frac{dy}{dx} = 4(3x^2 + 1)^3 \cdot 6x = 24x(3x^2 + 1)^3$
Answer: $24x(3x^2 + 1)^3$
Watch Out!
Remember that the derivative of a constant is always zero. Also, when using the product rule, be careful with the order: $f'g + fg'$, not $f'g'$.
Remember that the derivative of a constant is always zero. Also, when using the product rule, be careful with the order: $f'g + fg'$, not $f'g'$.
Practice for Differentiation Rules
- Find $\frac{dy}{dx}$ if $y = x^8$.
- Find $f'(x)$ if $f(x) = 5x^4 - 3x^2 + 7$.
- Find $\frac{dy}{dx}$ if $y = \frac{1}{x^3}$.
- Find $f'(x)$ if $f(x) = (2x - 5)(3x + 4)$ using the product rule.
- Find $\frac{dy}{dx}$ if $y = \frac{x^3}{x^2 + 1}$ using the quotient rule.
- Find $f'(x)$ if $f(x) = (x^2 - 4x)^5$ using the chain rule.
Gradient of a Curve at a Point
Example 1: Finding Gradient at a Point
Problem: Find the gradient of $y = x^3 - 2x^2 + 1$ at $x = 2$.
Formula: Gradient $m = f'(x)$ evaluated at the given point.
Solution:
$\frac{dy}{dx} = 3x^2 - 4x$
At $x = 2$: $\frac{dy}{dx} = 3(4) - 4(2) = 12 - 8 = 4$
Answer: The gradient is 4.
Problem: Find the gradient of $y = x^3 - 2x^2 + 1$ at $x = 2$.
Formula: Gradient $m = f'(x)$ evaluated at the given point.
Solution:
$\frac{dy}{dx} = 3x^2 - 4x$
At $x = 2$: $\frac{dy}{dx} = 3(4) - 4(2) = 12 - 8 = 4$
Answer: The gradient is 4.
Tangents and Normals
Example 1: Equation of a Tangent
Problem: Find the equation of the tangent to $y = x^2$ at $x = 3$.
Formula: $y - y_1 = m(x - x_1)$ where $m = f'(x_1)$
Solution:
$y = x^2$, $\frac{dy}{dx} = 2x$
At $x = 3$: $m = 2(3) = 6$, $y = 9$
Tangent: $y - 9 = 6(x - 3)$ → $y = 6x - 18 + 9 = 6x - 9$
Answer: $y = 6x - 9$
Problem: Find the equation of the tangent to $y = x^2$ at $x = 3$.
Formula: $y - y_1 = m(x - x_1)$ where $m = f'(x_1)$
Solution:
$y = x^2$, $\frac{dy}{dx} = 2x$
At $x = 3$: $m = 2(3) = 6$, $y = 9$
Tangent: $y - 9 = 6(x - 3)$ → $y = 6x - 18 + 9 = 6x - 9$
Answer: $y = 6x - 9$
Example 2: Equation of a Normal
Problem: Find the equation of the normal to $y = x^2$ at $x = 3$.
Formula: $m_{normal} = -\frac{1}{m_{tangent}}$
Solution:
From above, $m_{tangent} = 6$, so $m_{normal} = -\frac{1}{6}$
Normal: $y - 9 = -\frac{1}{6}(x - 3)$ → $y = -\frac{1}{6}x + \frac{1}{2} + 9 = -\frac{1}{6}x + \frac{19}{2}$
Answer: $y = -\frac{1}{6}x + \frac{19}{2}$
Problem: Find the equation of the normal to $y = x^2$ at $x = 3$.
Formula: $m_{normal} = -\frac{1}{m_{tangent}}$
Solution:
From above, $m_{tangent} = 6$, so $m_{normal} = -\frac{1}{6}$
Normal: $y - 9 = -\frac{1}{6}(x - 3)$ → $y = -\frac{1}{6}x + \frac{1}{2} + 9 = -\frac{1}{6}x + \frac{19}{2}$
Answer: $y = -\frac{1}{6}x + \frac{19}{2}$
Rate of Change
Applications of Rate of Change
Velocity: $v(t) = \frac{ds}{dt}$ where $s$ is displacement
Acceleration: $a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}$
Chain Rule for related rates: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$
Velocity: $v(t) = \frac{ds}{dt}$ where $s$ is displacement
Acceleration: $a(t) = \frac{dv}{dt} = \frac{d^2s}{dt^2}$
Chain Rule for related rates: $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$
Example 1: Velocity and Acceleration
Problem: The displacement of a particle is given by $s(t) = t^3 - 6t^2 + 9t$ metres. Find the velocity and acceleration at $t = 2$ seconds.
Formula: $v(t) = \frac{ds}{dt}$, $a(t) = \frac{dv}{dt}$
Solution:
$v(t) = \frac{ds}{dt} = 3t^2 - 12t + 9$
$a(t) = \frac{dv}{dt} = 6t - 12$
At $t = 2$: $v(2) = 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3$ m/s
$a(2) = 6(2) - 12 = 12 - 12 = 0$ m/s²
Answer: Velocity = -3 m/s, Acceleration = 0 m/s²
Problem: The displacement of a particle is given by $s(t) = t^3 - 6t^2 + 9t$ metres. Find the velocity and acceleration at $t = 2$ seconds.
Formula: $v(t) = \frac{ds}{dt}$, $a(t) = \frac{dv}{dt}$
Solution:
$v(t) = \frac{ds}{dt} = 3t^2 - 12t + 9$
$a(t) = \frac{dv}{dt} = 6t - 12$
At $t = 2$: $v(2) = 3(4) - 12(2) + 9 = 12 - 24 + 9 = -3$ m/s
$a(2) = 6(2) - 12 = 12 - 12 = 0$ m/s²
Answer: Velocity = -3 m/s, Acceleration = 0 m/s²
Example 2: Rate of Change of Area (Chain Rule)
Problem: The radius of a circle is increasing at 2 cm/s. Find the rate of change of area when $r = 5$ cm.
Formula: $\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}$
Solution:
Area $A = \pi r^2$, $\frac{dA}{dr} = 2\pi r$
$\frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt} = 2\pi r \cdot 2 = 4\pi r$
At $r = 5$: $\frac{dA}{dt} = 4\pi(5) = 20\pi$ cm²/s
Answer: $20\pi$ cm²/s
Problem: The radius of a circle is increasing at 2 cm/s. Find the rate of change of area when $r = 5$ cm.
Formula: $\frac{dA}{dt} = \frac{dA}{dr} \cdot \frac{dr}{dt}$
Solution:
Area $A = \pi r^2$, $\frac{dA}{dr} = 2\pi r$
$\frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt} = 2\pi r \cdot 2 = 4\pi r$
At $r = 5$: $\frac{dA}{dt} = 4\pi(5) = 20\pi$ cm²/s
Answer: $20\pi$ cm²/s
Cumulative Practice Exercises
- Differentiate $y = 4x^5 - 3x^3 + 2x^2 - 7$.
- Find $f'(x)$ if $f(x) = (3x^2 - 2x)(x^3 + 1)$ using the product rule.
- Find $\frac{dy}{dx}$ if $y = \frac{x^2 - 1}{x^2 + 1}$ using the quotient rule.
- Find $f'(x)$ if $f(x) = (2x^3 - 5x)^4$ using the chain rule.
- Find the gradient of $y = x^4 - 3x^2 + 2$ at $x = -1$.
- Find the equation of the tangent to $y = x^3 - 4x$ at $x = 2$.
- Find the equation of the normal to $y = x^2 - 3x + 2$ at $x = 1$.
- A particle moves with displacement $s(t) = t^3 - 9t^2 + 24t$. Find the velocity at $t = 3$.
- The radius of a sphere is increasing at 0.5 cm/s. Find the rate of change of volume when $r = 4$ cm. ($V = \frac{4}{3}\pi r^3$)
- Find the point on $y = x^2 - 2x + 3$ where the gradient is 2.
Answers to Cumulative Exercises
- Answer: $\frac{dy}{dx} = 20x^4 - 9x^2 + 4x$
- Answer: $f'(x) = 15x^4 - 8x^3 + 6x - 2$
- Answer: $\frac{dy}{dx} = \frac{4x}{(x^2+1)^2}$
- Answer: $f'(x) = 4(2x^3 - 5x)^3(6x^2 - 5)$
- Answer: Gradient = 2
- Answer: $y = 8x - 16$
- Answer: $y = x - 1$
- Answer: $v(3) = -3$ m/s
- Answer: $\frac{dV}{dt} = 32\pi$ cm³/s
- Answer: Point $(2, 3)$
Conclusion & Summary
Differentiation is a fundamental tool in calculus that allows us to find gradients, equations of tangents and normals, and rates of change. The derivative $f'(x)$ represents the instantaneous rate of change and the slope of the tangent line at any point. Mastery of differentiation rules—power rule, product rule, quotient rule, and chain rule—enables us to differentiate a wide variety of functions.
Key Takeaways:
1. First Principles: $f'(x) = \lim_{\Delta x \to 0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$
2. Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$
3. Product Rule: $(fg)' = f'g + fg'$
4. Quotient Rule: $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$
5. Chain Rule: $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$
6. Tangent: $y - y_0 = m(x - a)$ where $m = f'(a)$
7. Normal: $y - y_0 = -\frac{1}{m}(x - a)$ (if $m \neq 0$)
8. Rate of Change: $\frac{dy}{dx}$ is the instantaneous rate of change.
Keep practising differentiation rules and applications. Calculus is the language of change in science and engineering!
Video Resource
Watch this video for more examples of differentiation, tangents, normals, and rate of change problems.
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