Differentiation

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1. Standard Derivatives

What is differentiation? Differentiation is a way of finding how fast something is changing. The derivative (written as dy/dx or f'(x)) tells us the rate of change or the gradient (slope) of a function at any point.

Important derivatives you need to know:

These are formulas you should memorize:

• If y = e^x, then dy/dx = e^x
• If y = ln x, then dy/dx = 1/x
• If y = sin x, then dy/dx = cos x
• If y = cos x, then dy/dx = -sin x
• If y = tan x, then dy/dx = sec²x (which is the same as 1/cos²x)

What does this mean? The derivative of e^x is just e^x itself - that's a special property! For ln x, the derivative is 1/x. For the trigonometric functions (sin, cos, tan), each has its own derivative pattern.

2. Constant Multiples, Sums, and Differences

Constant multiples: If you have a constant (a number) multiplied by a function, you can bring the constant outside when differentiating.

Rule: If y = cf(x), then dy/dx = c × f'(x)

Example:

• If y = 5e^x, then dy/dx = 5e^x
• If y = 3 sin x, then dy/dx = 3 cos x
• If y = 4 ln x, then dy/dx = 4/x

Sums and differences: When you have functions added or subtracted, you differentiate each part separately.

Rule: If y = f(x) + g(x), then dy/dx = f'(x) + g'(x)

Example: If y = e^x + sin x, then dy/dx = e^x + cos x

If y = 3 ln x - 2 cos x, then dy/dx = 3/x - 2(-sin x) = 3/x + 2 sin x

3. The Chain Rule (Composite Functions)

What is a composite function? A composite function is a "function of a function" - one function is inside another. For example, sin(3x) has 3x inside the sin function.

The Chain Rule: When differentiating a composite function, you multiply the derivative of the outer function by the derivative of the inner function.

Formula: If y = f(g(x)), then dy/dx = f'(g(x)) × g'(x)

In simple terms: Differentiate the outside, leave the inside alone, then multiply by the derivative of the inside.

Example 1: y = e^(2x)

• Outside function: e^(something)
• Inside function: 2x
• Derivative of outside: e^(2x) (leave the inside as it is)
• Derivative of inside: 2
• Answer: dy/dx = e^(2x) × 2 = 2e^(2x)

Example 2: y = ln(3x + 1)

• Outside function: ln(something)
• Inside function: 3x + 1
• Derivative of outside: 1/(3x + 1)
• Derivative of inside: 3
• Answer: dy/dx = 1/(3x + 1) × 3 = 3/(3x + 1)

Example 3: y = sin(x²)

• Outside function: sin(something)
• Inside function: x²
• Derivative of outside: cos(x²)
• Derivative of inside: 2x
• Answer: dy/dx = cos(x²) × 2x = 2x cos(x²)

Example 4: y = (2x - 3)⁵

• Outside function: (something)⁵
• Inside function: 2x - 3
• Derivative of outside: 5(2x - 3)⁴
• Derivative of inside: 2
• Answer: dy/dx = 5(2x - 3)⁴ × 2 = 10(2x - 3)⁴

4. The Product Rule

When do you use it? Use the product rule when you have two functions multiplied together.

The Product Rule: If y = u × v (where u and v are both functions of x), then:

dy/dx = u(dv/dx) + v(du/dx)

In words: First function × derivative of second + second function × derivative of first

Example 1: y = x² ln x

• Let u = x², so du/dx = 2x
• Let v = ln x, so dv/dx = 1/x
• Using the product rule: dy/dx = x² × (1/x) + ln x × 2x dy/dx = x + 2x ln x

Example 2: y = xe^(1-x²)

• Let u = x, so du/dx = 1
• Let v = e^(1-x²), so dv/dx = e^(1-x²) × (-2x) = -2xe^(1-x²) (using chain rule)
• Using the product rule: dy/dx = x × (-2xe^(1-x²)) + e^(1-x²) × 1 dy/dx = -2x²e^(1-x²) + e^(1-x²) dy/dx = e^(1-x²)(1 - 2x²)

Example 3: y = e^x sin x

• Let u = e^x, so du/dx = e^x
• Let v = sin x, so dv/dx = cos x
• Using the product rule: dy/dx = e^x × cos x + sin x × e^x dy/dx = e^x(cos x + sin x)

5. The Quotient Rule

When do you use it? Use the quotient rule when you have one function divided by another function (a fraction with x in both the top and bottom).

The Quotient Rule: If y = u/v (where u and v are both functions of x), then:

dy/dx = [v(du/dx) - u(dv/dx)] / v²

In words: (Bottom × derivative of top - top × derivative of bottom) / (bottom)²

Memory tip: Remember "lo d-hi minus hi d-lo over lo-lo" (lo = lower/bottom, hi = higher/top, d = derivative)

Example 1: y = (2x - 4)/(3x + 2)

• Let u = 2x - 4, so du/dx = 2
• Let v = 3x + 2, so dv/dx = 3
• Using the quotient rule: dy/dx = [(3x + 2)(2) - (2x - 4)(3)] / (3x + 2)² dy/dx = [6x + 4 - 6x + 12] / (3x + 2)² dy/dx = 16 / (3x + 2)²

Example 2: y = x/ln x

• Let u = x, so du/dx = 1
• Let v = ln x, so dv/dx = 1/x
• Using the quotient rule: dy/dx = [ln x × 1 - x × (1/x)] / (ln x)² dy/dx = [ln x - 1] / (ln x)²

Example 3: y = e^x / (x + 1)

• Let u = e^x, so du/dx = e^x
• Let v = x + 1, so dv/dx = 1
• Using the quotient rule: dy/dx = [(x + 1)e^x - e^x × 1] / (x + 1)² dy/dx = [xe^x + e^x - e^x] / (x + 1)² dy/dx = xe^x / (x + 1)²

6. Parametric Differentiation

What are parametric equations? Sometimes, instead of y being directly in terms of x, both x and y are given in terms of a third variable (usually called t). This third variable is called a parameter.

Example of parametric equations:

• x = t - e^(2t)
• y = t + e^(2t)

Here, both x and y depend on t, not directly on each other.

How to find dy/dx: To find the gradient dy/dx when you have parametric equations, use this formula:

dy/dx = (dy/dt) / (dx/dt)

In words: Differentiate y with respect to t, then divide by the derivative of x with respect to t.

Step-by-step method:

1. Differentiate x with respect to t to get dx/dt
2. Differentiate y with respect to t to get dy/dt
3. Divide: dy/dx = (dy/dt) ÷ (dx/dt)

Example: Find dy/dx when x = t - e^(2t) and y = t + e^(2t)

Step 1: Find dx/dt

• x = t - e^(2t)
• dx/dt = 1 - 2e^(2t) (using chain rule on e^(2t))

Step 2: Find dy/dt

• y = t + e^(2t)
• dy/dt = 1 + 2e^(2t)

Step 3: Find dy/dx

• dy/dx = (dy/dt) / (dx/dt)
• dy/dx = (1 + 2e^(2t)) / (1 - 2e^(2t))

Another example: x = 2t², y = 4t

Step 1: dx/dt = 4t Step 2: dy/dt = 4 Step 3: dy/dx = 4/(4t) = 1/t

7. Implicit Differentiation

What is an implicit function? Usually, equations are written with y = something (this is called explicit form). But sometimes, x and y are mixed together in the equation and you can't easily separate them. This is called an implicit equation.

Example of implicit equations:

• x² + y² = xy + 7
• x³ + y³ = 6xy

How to differentiate implicitly: The key idea is to differentiate both sides of the equation with respect to x, remembering that y is a function of x.

Important rule: When differentiating any term with y, you must multiply by dy/dx because y depends on x.

Examples of implicit differentiation:

• d/dx(y) = dy/dx
• d/dx(y²) = 2y × dy/dx (using chain rule)
• d/dx(y³) = 3y² × dy/dx
• d/dx(xy) = x × dy/dx + y × 1 (using product rule)

Step-by-step method:

1. Differentiate both sides of the equation with respect to x
2. Remember to multiply by dy/dx whenever you differentiate a term containing y
3. Collect all terms with dy/dx on one side
4. Factor out dy/dx
5. Solve for dy/dx

Example 1: Find dy/dx when x² + y² = xy + 7

Step 1: Differentiate both sides with respect to x

• Left side: d/dx(x²) + d/dx(y²) = 2x + 2y(dy/dx)
• Right side: d/dx(xy) + d/dx(7) = x(dy/dx) + y + 0

Step 2: Write the equation

• 2x + 2y(dy/dx) = x(dy/dx) + y

Step 3: Collect dy/dx terms on one side

• 2y(dy/dx) - x(dy/dx) = y - 2x

Step 4: Factor out dy/dx

• dy/dx(2y - x) = y - 2x

Step 5: Solve for dy/dx

• dy/dx = (y - 2x)/(2y - x)

Example 2: Find dy/dx when x³ + y³ = 6xy

Step 1: Differentiate both sides

• 3x² + 3y²(dy/dx) = 6[x(dy/dx) + y]
• 3x² + 3y²(dy/dx) = 6x(dy/dx) + 6y

Step 2: Collect dy/dx terms

• 3y²(dy/dx) - 6x(dy/dx) = 6y - 3x²

Step 3: Factor

• dy/dx(3y² - 6x) = 6y - 3x²

Step 4: Solve

• dy/dx = (6y - 3x²)/(3y² - 6x)
• dy/dx = (2y - x²)/(y² - 2x) (after dividing by 3)

8. Tangents and Normals

What are tangents and normals?

• A tangent is a straight line that just touches a curve at one point. It has the same gradient as the curve at that point.
• A normal is a straight line perpendicular (at right angles) to the tangent at that point.

Finding the equation of a tangent:

Step 1: Find the value of dy/dx at the given point. This is the gradient of the tangent.

Step 2: Use the equation of a straight line: y - y₁ = m(x - x₁)

• Where m is the gradient (dy/dx)
• (x₁, y₁) is the point on the curve

Finding the equation of a normal:

Step 1: Find the gradient of the tangent (dy/dx at the point)

Step 2: The gradient of the normal is the negative reciprocal of the tangent's gradient.

• If tangent gradient = m, then normal gradient = -1/m

Step 3: Use y - y₁ = m_normal(x - x₁)

Example with explicit function: Find the equation of the tangent and normal to the curve y = x² + 3x at the point (1, 4).

For the tangent:

• Step 1: dy/dx = 2x + 3 At x = 1: dy/dx = 2(1) + 3 = 5 (gradient of tangent)
• Step 2: Using y - y₁ = m(x - x₁) y - 4 = 5(x - 1) y - 4 = 5x - 5 y = 5x - 1 (equation of tangent)

For the normal:

• Gradient of normal = -1/5
• y - 4 = (-1/5)(x - 1)
• y - 4 = -x/5 + 1/5
• 5y - 20 = -x + 1
• 5y = -x + 21 or x + 5y = 21 (equation of normal)

Example with parametric equations: Find the equation of the tangent to the curve x = t², y = 2t at the point where t = 1.

Step 1: Find the coordinates at t = 1

• x = 1² = 1
• y = 2(1) = 2
• Point is (1, 2)

Step 2: Find dy/dx

• dx/dt = 2t
• dy/dt = 2
• dy/dx = 2/(2t) = 1/t
• At t = 1: dy/dx = 1

Step 3: Equation of tangent

• y - 2 = 1(x - 1)
• y = x + 1

Example with implicit function: Find the gradient of the tangent to x² + y² = 25 at the point (3, 4).

Step 1: Differentiate implicitly

• 2x + 2y(dy/dx) = 0
• dy/dx = -2x/(2y) = -x/y

Step 2: At point (3, 4)

• dy/dx = -3/4

Step 3: Equation of tangent

• y - 4 = (-3/4)(x - 3)
• 4y - 16 = -3x + 9
• 4y = -3x + 25 or 3x + 4y = 25