P1: An Introduction to Differentiation

The Magic of Slopes & Turning Points ✨

Introduction

1. Introduction

Welcome to Differentiation! Ever wondered how to find the exact gradient of a curvy line at a single point? That's the core idea we're about to master. We'll kick things off by learning the fundamental power rule, your new best friend for differentiating any polynomial. Then, we'll use that skill to find the equations of tangents and normals to curves - basically, lines that just touch or are perpendicular to a graph at a specific point. The real power move comes next: locating and classifying stationary points . This is how you'll find the exact coordinates of maximums, minimums, and other key features of a function. Finally, we'll dive into connected rates of change, which sounds intense but is just a cool way to see how the rate of change of one variable affects another, like how fast a balloon's radius grows as it's inflated. Let's get started!

x^n

y=(3x+5)^4

Worked example

Let's Crack This Problem 🤓

y = \sqrt{6x-2}

(x_1, y_1)

m_T

Worked example

Let's put this theory into practice! 🚀

y = 2x^2 - 8x + 3

\frac{dy}{dx}

\frac{d^2y}{dx^2}

Worked example

Let's put this theory into practice! 🚀

y = 2x^3 - 3x^2 - 12x + 4

1
$\frac{dy}{dx}$
$\frac{dy}{dx} = 6x^2 - 6x - 12$
2
$x$
$6x^2 - 6x - 12 = 0 \\ x^2 - x - 2 = 0 \\ (x-2)(x+1) = 0 \\ x = 2 \text{ or } x = -1$
3
$y$
$\text{When } x=2, y = 2(2)^3 - 3(2)^2 - 12(2) + 4 = -16 \implies \text{Point is } (2, -16)$
4
$\frac{d^2y}{dx^2}$
$\frac{d^2y}{dx^2} = 12x - 6$
5
$Now for the final verdict. We substitute each x-coordinate into the second derivative. A positive result means it's a minimum; a negative result means it's a maximum.$
$\text{At } x=2: \frac{d^2y}{dx^2} = 12(2) - 6 = 18. \text{ Since } 18 > 0, \text{ the point } (2, -16) \text{ is a local minimum.}$

Answer

\text{At } x=2: \frac{d^2y}{dx^2} = 12(2) - 6 = 18. \text{ Since } 18 > 0, \text{ the point } (2, -16) \text{ is a local minimum.}

y

dx

Worked example

Let's blow this problem up! 🎈

50 \text{ cm}^3\text{s}^{-1}

1
$First, let's identify what we know and what we need to find. We're given the rate of change of volume with respect to time, and we need to find the rate of change of the radius with respect to time.$
$\text{Given: } \frac{dV}{dt} = 50 \text{ cm}^3\text{s}^{-1} <br> \text{Find: } \frac{dr}{dt} \text{ when } r = 10 \text{ cm}$
2
$V$
$V = \frac{4}{3}\pi r^3 <br> \frac{dV}{dr} = \frac{4}{3}\pi \times 3r^2 = 4\pi r^2$
3
$V$
$\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt}$
4
$\frac{dr}{dt}$
$\frac{dr}{dt} = \frac{dV}{dt} \div \frac{dV}{dr} = \frac{\frac{dV}{dt}}{\frac{dV}{dr}} <br> \text{When } r=10, \quad \frac{dV}{dr} = 4\pi (10)^2 = 400\pi$
5
$\frac{dV}{dt}$
$\frac{dr}{dt} = \frac{50}{400\pi} = \frac{1}{8\pi} <br> \text{So, the radius is increasing at a rate of } \frac{1}{8\pi} \text{ cm s}^{-1}.$

Answer

\frac{dr}{dt} = \frac{50}{400\pi} = \frac{1}{8\pi} <br> \text{So, the radius is increasing at a rate of } \frac{1}{8\pi} \text{ cm s}^{-1}.

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