Chapter 3: Advanced Differentiation Techniques

Unlocking Calculus Superpowers! 🚀

Introduction

1. Introduction

y = f(x)

2. Derivatives of Standard Functions and the Chain Rule

e^x

$A summary table showing the standard function (e.g., e^x, ln x, sin x, cos x, tan x, tan⁻¹x) in one column and its derivative in the next column. A third column shows the chain rule version, e.g., for e^{f(x)}, the derivative is f'(x)e^{f(x)}.$

It's a pattern, and once you see it, you can apply it to any combination they throw at you. You've got this!

Worked example

Worked Example: Chain Rule with Exponential and Trigonometric Functions

Let's Solve This Thing 🚀

y = 5e^{\sin(2x)}

1
$5e^u$
$y = 5e^{\sin(2x)} \implies \text{Outer: } 5e^u, \text{ Middle: } \sin(v), \text{ Inner: } 2x$
2
$5e^u$
$\frac{d}{du}(5e^u) = 5e^u = 5e^{\sin(2x)}$
3
$u = \sin(v)$
$\frac{du}{dv}(\sin(v)) = \cos(v) = \cos(2x)$
4
$v = 2x$
$\frac{dv}{dx}(2x) = 2$
5
$Now for the magic! According to the chain rule, we multiply all these results together to get our final answer. Think of it as `(Derivative of Outer) (Derivative of Middle) (Derivative of Inner)`.$
$\frac{dy}{dx} = (5e^{\sin(2x)}) \times (\cos(2x)) \times (2)$
6
$Let's clean it up. We just need to rearrange the terms to write the final answer in the standard format, usually with the constant and polynomial/trig parts at the front.$
$\frac{dy}{dx} = 10\cos(2x)e^{\sin(2x)}$

Answer

\frac{dy}{dx} = 10\cos(2x)e^{\sin(2x)}

3. Differentiation of Products and Quotients

u

A colourful diagram of the Product Rule formula, showing u and v as two distinct blocks. Arrows point from v to dv/dx and from u to du/dx, then combine into the final formula.

y = \frac{u}{v}

Worked example

Worked Example: Applying the Product Rule

Let's Differentiate This Combo! 🎮

y = x^2 e^{5x}

4. Parametric and Implicit Differentiation

y

A diagram showing a parametric curve, like a spiral, with an arrow indicating the direction of increasing 't'. A point P on the curve is marked, with a tangent line drawn through it.

x

An image of an implicitly defined curve, like an ellipse (e.g., 4x^2 + 9y^2 = 36). A point (x, y) on the curve is shown with its tangent line, demonstrating that the gradient can be found at any point.

Worked example

Worked Example: Tangent to a Parametric Curve

Let's Find That Gradient! 🚀

x = 2t + \ln(t)

1
$x$
$\frac{dx}{dt} = 2 + \frac{1}{t} \quad \text{and} \quad \frac{dy}{dt} = 2t$
2
$\frac{dy}{dx}$
$\frac{dy}{dx} = \frac{2t}{2 + \frac{1}{t}} = \frac{2t}{\frac{2t+1}{t}} = \frac{2t^2}{2t+1}$
3
$t=1$
$m = \frac{2(1)^2}{2(1)+1} = \frac{2}{3}$
4
$(x, y)$
$x = 2(1) + \ln(1) = 2 + 0 = 2 \\ y = (1)^2 - 1 = 1 - 1 = 0 \\ \text{So, the point is } (2, 0).$
5
$(2, 0)$
$y - 0 = \frac{2}{3}(x - 2) \implies y = \frac{2}{3}x - \frac{4}{3} \\ \text{Or, in a cleaner form: } 3y - 2x + 4 = 0$

Answer

y - 0 = \frac{2}{3}(x - 2) \implies y = \frac{2}{3}x - \frac{4}{3} \\ \text{Or, in a cleaner form: } 3y - 2x + 4 = 0

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