Finding Slope at a Point - Derivative Evaluation Guide

The derivative gives you a formula for slope. But most exam questions don't stop there — they ask for the slope at a specific point, or the equation of the tangent line, or whether a function is increasing or decreasing.

Here's how to go from f'(x) to a concrete number.

The Two-Step Process

1. Find the derivative f'(x) using differentiation rules
2. Plug in the x-value to get the slope at that point

Example: f(x) = 2x³ − 5x² + 3x − 1 at x = 2

Step 1: f'(x) = 6x² − 10x + 3

Step 2: f'(2) = 6(4) − 10(2) + 3 = 24 − 20 + 3 = 7

The slope at x = 2 is 7. The curve is rising at that point.

Finding the Tangent Line Equation

A tangent line needs a point and a slope. You have both:

• Point: (2, f(2)) = (2, 2(8) − 5(4) + 3(2) − 1) = (2, 16 − 20 + 6 − 1) = (2, 1)
• Slope: f'(2) = 7

Using point-slope form: y − 1 = 7(x − 2) → y = 7x − 13

This line touches the curve at exactly one point and has the same slope as the curve at that point.

Application: Velocity from Position

If s(t) gives position at time t, then v(t) = s'(t) gives velocity.

Example: A ball's height is s(t) = −16t² + 48t + 5 (feet, seconds)

• v(t) = s'(t) = −32t + 48
• v(1) = −32 + 48 = 16 ft/s (moving up)
• v(1.5) = −48 + 48 = 0 ft/s (peak — ball stops momentarily)
• v(2) = −64 + 48 = −16 ft/s (falling down)

The derivative tells you direction: positive = up, zero = peak, negative = down.

Application: Marginal Cost in Economics

If C(x) is the total cost of producing x units, then C'(x) = marginal cost — the cost of producing one additional unit.

Example: C(x) = 0.01x³ − 0.3x² + 50x + 1000

• C'(x) = 0.03x² − 0.6x + 50
• C'(100) = 300 − 60 + 50 = ₹290 per additional unit
• C'(200) = 1200 − 120 + 50 = ₹1,130 per additional unit

Marginal cost increases as production scales — a real phenomenon in manufacturing.

Determining Where a Function Increases/Decreases

| Condition | Behavior | |-----------|----------| | f'(x) > 0 | Function is increasing | | f'(x) < 0 | Function is decreasing | | f'(x) = 0 | Critical point (potential max/min) |

Example: f(x) = x³ − 3x

f'(x) = 3x² − 3 = 3(x² − 1) = 3(x+1)(x−1)

Critical points: x = −1 and x = 1

| Interval | Test x | f'(x) sign | Behavior | |----------|--------|------------|----------| | x < −1 | −2 | 3(4−1) = +9 | Increasing | | −1 < x < 1 | 0 | 3(0−1) = −3 | Decreasing | | x > 1 | 2 | 3(4−1) = +9 | Increasing |

So x = −1 is a local max and x = 1 is a local min.

The Trench Truth: When a question asks "find where f is increasing," it's asking for the x-intervals where f'(x) > 0 — NOT where f(x) > 0. Students confuse the function with its derivative constantly. Read the question twice.

Try it yourself with our derivative calculator — enter a function and an evaluation point to get the exact slope instantly.

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