Further Calculus | Highers - Wyatt's Notes

Further Differentiation

Implicit Differentiation

When a function is defined implicitly (e.g., $x^2 + y^2 = 25$ ), differentiate both sides with Respect to $x$ Treating $y$ as a function of $x$ .

\frac{d}{dx}[y^n] = ny^{n-1}\frac{dy}{dx}

This is the chain rule applied to $y(x)^n$ : $\frac{d}{dx}[y^n] = ny^{n-1} \cdot y"(x)$ .

Example: Find $\dfrac{dy}{dx}$ for $x^2 + y^2 = 25$ .

2x + 2y\frac{dy}{dx} = 0

\frac{dy}{dx} = -\frac{x}{y}

This makes geometric sense: on the upper semicircle ( $y > 0$ ), increasing $x$ decreases $y$ Giving A negative slope. On the lower semicircle ( $y < 0$ ), the slope is positive.

Example: Find $\dfrac{dy}{dx}$ for $x^3 + y^3 = 6xy$ .

Differentiate implicitly:

3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}

3y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2

\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2

\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}

Example: Find the tangent to $x^2 + y^2 + 2x - 4y = 11$ at the point $(1, 2)$ .

Verify: $1 + 4 + 2 - 8 = -1 \neq 11$ . The point is not on the curve. Let us find the gradient Function first:

2x + 2y\frac{dy}{dx} + 2 - 4\frac{dy}{dx} = 0

(2y - 4)\frac{dy}{dx} = -2x - 2

\frac{dy}{dx} = \frac{-2x - 2}{2y - 4} = \frac{-x - 1}{y - 2}

Second derivatives implicitly. Differentiate $\frac{dy}{dx}$ again with respect to $x$ Using The chain rule wherever $y$ appears.

Parametric Differentiation

If $x = f(t)$ and $y = g(t)$ Then:

\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}

This follows from the chain rule: $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ .

Second Derivative:

\frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{dx/dt}

Example: A curve has parametric equations $x = 2\cos t$ , $y = \sin 2t$ . Find $\dfrac{dy}{dx}$ at $t = \dfrac{\pi}{4}$ .

\frac{dx}{dt} = -2\sin t, \quad \frac{dy}{dt} = 2\cos 2t

\frac{dy}{dx} = \frac{2\cos 2t}{-2\sin t} = -\frac{\cos 2t}{\sin t}

At $t = \dfrac{\pi}{4}$ :

\frac{dy}{dx} = -\frac{\cos(\pi/2)}{\sin(\pi/4)} = -\frac{0}{1/\sqrt{2}} = 0

Example: Find $\frac{d^2y}{dx^2}$ for $x = t^2$ , $y = t^3$ .

\frac{dy}{dx} = \frac{3t^2}{2t} = \frac{3t}{2}

\frac{d^2y}{dx^2} = \frac{d/dt(3t/2)}{dx/dt} = \frac{3/2}{2t} = \frac{3}{4t}

Differentiation of Inverse Trigonometric Functions

\frac{d}{dx}[\arcsin x] = \frac{1}{\sqrt{1 - x^2}}, \quad |x| < 1

\frac{d}{dx}[\arccos x] = \frac{-1}{\sqrt{1 - x^2}}, \quad |x| < 1

\frac{d}{dx}[\arctan x] = \frac{1}{1 + x^2}

Proof that $\frac{d}{dx}[\arctan x] = \frac{1}{1+x^2}$ . Let $y = \arctan x$ So $x = \tan y$ . Differentiating implicitly: $1 = \sec^2 y \cdot \frac{dy}{dx}$ Giving $\frac{dy}{dx} = \frac{1}{\sec^2 y} = \frac{1}{1 + \tan^2 y} = \frac{1}{1 + x^2}$ .

Example: Differentiate $f(x) = \arcsin(3x)$ .

F'(x) = \frac{1}{\sqrt{1 - (3x)^2}} \cdot 3 = \frac{3}{\sqrt{1 - 9x^2}}

Logarithmic Differentiation

For functions of the form $y = [f(x)]^{g(x)}$ Take natural logarithms first:

\ln y = g(x) \ln f(x)

Then differentiate implicitly.

Example: Differentiate $y = x^x$ .

\ln y = x \ln x

\frac{1}{y}\frac{dy}{dx} = \ln x + x \cdot \frac{1}{x} = \ln x + 1

\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)

Example: Differentiate $y = x^{\sin x}$ .

\ln y = \sin x \cdot \ln x

\frac{1}{y}\frac{dy}{dx} = \cos x \cdot \ln x + \sin x \cdot \frac{1}{x}

\frac{dy}{dx} = x^{\sin x}\!\left(\cos x \ln x + \frac{\sin x}{x}\right)

Related rates problems involve finding the rate of change of one quantity given the rate of change Of a related quantity.

Example: A ladder 5 m long leans against a wall. The bottom slides away at 0.5 m/s. How fast is The top sliding down when the bottom is 3 m from the wall?

Let $x$ be the distance from the wall and $y$ be the height. Then $x^2 + y^2 = 25$ .

Differentiate: $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ .

When $x = 3$ : $y = \sqrt{25 - 9} = 4$ .

2(3)(0.5) + 2(4)\frac{dy}{dt} = 0 \implies 3 + 8\frac{dy}{dt} = 0 \implies \frac{dy}{dt} = -\frac{3}{8} \mathrm{ m/s

Further Integration

Integration by Parts

\int u \, dv = uv - \int v \, du

Proof. By the product rule: $\frac{d}{dx}[uv] = u'v + uv'$ So $uv' = \frac{d}{dx}[uv] - u'v$ . Integrating both sides: $\int uv'\,dx = uv - \int u'v\,dx$ .

LIATE rule for choosing $u$ (in order of priority): Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Choose $u$ as the function that appears earliest in this list.

Example: Evaluate $\int x e^{2x} \, dx$ .

Let $u = x$$dv = e^{2x} dx$ . Then $du = dx$$v = \dfrac{e^{2x}}{2}$ .

\int x e^{2x} \, dx = x \cdot \frac{e^{2x}}{2} - \int \frac{e^{2x}}{2} \, dx = \frac{x e^{2x}}{2} - \frac{e^{2x}}{4} + C = \frac{e^{2x}(2x - 1)}{4} + C

Example: Evaluate $\int x^2 \cos x \, dx$ .

Let $u = x^2$$dv = \cos x \, dx$ . Then $du = 2x \, dx$$v = \sin x$ .

= x^2 \sin x - \int 2x \sin x \, dx

Apply integration by parts again for $\int x \sin x \, dx$ :

Let $u = x$$dv = \sin x \, dx$ . Then $du = dx$$v = -\cos x$ .

\int x \sin x \, dx = -x\cos x + \int \cos x \, dx = -x\cos x + \sin x

So:

\int x^2 \cos x \, dx = x^2 \sin x - 2(-x\cos x + \sin x) + C = x^2 \sin x + 2x\cos x - 2\sin x + C

Cyclic Integration by Parts

Some integrals require integration by parts twice, then solving algebraically.

Example: Evaluate $\int e^x \cos x \, dx$ .

Let $u = e^x$$dv = \cos x\,dx$ . Then $du = e^x\,dx$$v = \sin x$ .

I = e^x \sin x - \int e^x \sin x\,dx

Apply parts again for $\int e^x \sin x\,dx$ : $u = e^x$$dv = \sin x\,dx$ .

\int e^x \sin x\,dx = -e^x \cos x + \int e^x \cos x\,dx = -e^x \cos x + I

Substituting back:

I = e^x \sin x - (-e^x \cos x + I) = e^x \sin x + e^x \cos x - I

2I = e^x(\sin x + \cos x)

I = \frac{e^x(\sin x + \cos x)}{2} + C

Integration by Substitution

\int f(g(x))g'(x) \, dx = \int f(u) \, du \quad \mathrm{where u = g(x)

Example: Evaluate $\int \dfrac{x}{x^2 + 1} \, dx$ .

Let $u = x^2 + 1$ So $du = 2x \, dx$ Giving $\dfrac{1}{2} du = x \, dx$ .

\int \frac{x}{x^2 + 1} \, dx = \frac{1}{2}\int \frac{du}{u} = \frac{1}{2}\ln|u| + C = \frac{1}{2}\ln(x^2 + 1) + C

Definite Integration by Substitution

When using substitution in a definite integral, you must change the limits.

Example: Evaluate $\displaystyle\int_0^1 \frac{2x}{x^2 + 1}\,dx$ .

Let $u = x^2 + 1$ . When $x = 0$$u = 1$ . When $x = 1$$u = 2$ .

\int_0^1 \frac{2x}{x^2 + 1}\,dx = \int_1^2 \frac{du}{u} = [\ln u]_1^2 = \ln 2 - \ln 1 = \ln 2

Partial Fractions in Integration

Rational functions can be decomposed using partial fractions to make them easier to integrate.

Example: Evaluate $\displaystyle\int \frac{2x + 1}{(x + 1)(x - 2)} \, dx$ .

\frac{2x + 1}{(x + 1)(x - 2)} = \frac{A}{x + 1} + \frac{B}{x - 2}

2x + 1 = A(x - 2) + B(x + 1)

$x = -1$ : $-2 + 1 = A(-3)$ So $A = \dfrac{1}{3}$ .

$x = 2$ : $4 + 1 = B(3)$ So $B = \dfrac{5}{3}$ .

\int \frac{2x + 1}{(x + 1)(x - 2)} \, dx = \frac{1}{3}\int \frac{dx}{x + 1} + \frac{5}{3}\int \frac{dx}{x - 2} = \frac{1}{3}\ln|x + 1| + \frac{5}{3}\ln|x - 2| + C

Integration of Trigonometric Functions

\int \tan x \, dx = -\ln|\cos x| + C = \ln|\sec x| + C

\int \cot x \, dx = \ln|\sin x| + C

\int \sec x \, dx = \ln|\sec x + \tan x| + C

Example: Evaluate $\int \tan^2 x \, dx$ .

\int \tan^2 x \, dx = \int (\sec^2 x - 1) \, dx = \tan x - x + C

Example: Evaluate $\displaystyle\int \frac{dx}{\sin^2 x + 4\cos^2 x}$ .

Divide numerator and denominator by $\cos^2 x$ :

= \int \frac{\sec^2 x}{\tan^2 x + 4} \, dx

Let $u = \tan x$$du = \sec^2 x \, dx$ :

= \int \frac{du}{u^2 + 4} = \frac{1}{2}\arctan\left(\frac{u}{2}\right) + C = \frac{1}{2}\arctan\left(\frac{\tan x}{2}\right) + C

Volumes of Revolution

The volume generated by rotating $y = f(x)$ about the $x$ -axis from $x = a$ to $x = b$ :

V = \pi \int_a^b [f(x)]^2 \, dx

The volume generated by rotating about the $y$ -axis:

V = \pi \int_c^d [f^{-1}(y)]^2 \, dy

Example: Find the volume generated by rotating $y = \sqrt{x}$ about the $x$ -axis from $x = 0$ to $x = 4$ .

V = \pi \int_0^4 x \, dx = \pi\left[\frac{x^2}{2}\right]_0^4 = 8\pi

Differential Equations

First Order Separable Equations

Separable Equations: Equations of the form $\dfrac{dy}{dx} = f(x)g(y)$ can be solved by Separating variables:

\int \frac{dy}{g(y)} = \int f(x) \, dx

Example: Solve $\dfrac{dy}{dx} = \dfrac{x}{y}$ with $y(0) = 2$ .

Y \, dy = x \, dx

\int y \, dy = \int x \, dx

\frac{y^2}{2} = \frac{x^2}{2} + C

Y^2 = x^2 + 2C

Using $y(0) = 2$ : $4 = 0 + 2C$ So $C = 2$ .

Y^2 = x^2 + 4

Since $y(0) = 2 > 0$ : $y = \sqrt{x^2 + 4}$ .

First Order Linear Differential Equations

Equations of the form:

\frac{dy}{dx} + P(x)y = Q(x)

Integrating Factor: $\mu(x) = e^{\int P(x) \, dx}$

Multiply through by the integrating factor:

\frac{d}{dx}[\mu y] = \mu Q

Y = \frac{1}{\mu}\int \mu Q \, dx

Why this works. We want to write the left side as the derivative of a product. If we multiply by $\mu$ :

\mu\frac{dy}{dx} + \mu P y = \mu Q

This is $\frac{d}{dx}[\mu y]$ provided $\mu' = \mu P$ I.e., $\mu = e^{\int P\,dx}$ .

Example: Solve $\dfrac{dy}{dx} + 3y = 6e^{-3x}$ .

Integrating factor: $\mu = e^{\int 3 \, dx} = e^{3x}$ .

E^{3x}\frac{dy}{dx} + 3e^{3x}y = 6

\frac{d}{dx}[e^{3x}y] = 6

E^{3x}y = 6x + C

Y = e^{-3x}(6x + C)

Second Order Differential Equations

Homogeneous equations with constant coefficients:

A\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = 0

Try $y = e^{mx}$ : the auxiliary equation is $am^2 + bm + c = 0$ .

Case 1: Two distinct real roots $m_1, m_2$ :

Y = Ae^{m_1 x} + Be^{m_2 x}

Case 2: Repeated root $m$ :

Y = (Ax + B)e^{mx}

Case 3: Complex roots $m = \alpha \pm \beta i$ :

Y = e^{\alpha x}(A\cos\beta x + B\sin\beta x)

Why the complex case produces sines and cosines. If $m = \alpha + \beta i$ Then $e^{(\alpha+\beta i)x} = e^{\alpha x}(\cos\beta x + i\sin\beta x)$ . Since the original DE has real Coefficients, both $e^{(\alpha+\beta i)x}$ and $e^{(\alpha-\beta i)x}$ are solutions. Their linear Combinations give $e^{\alpha x}\cos\beta x$ and $e^{\alpha x}\sin\beta x$ .

Example: Solve $\dfrac{d^2y}{dx^2} - 5\dfrac{dy}{dx} + 6y = 0$ .

Auxiliary equation: $m^2 - 5m + 6 = 0$ So $(m - 2)(m - 3) = 0$ Giving $m = 2, 3$ .

Y = Ae^{2x} + Be^{3x}

Example: Solve $\dfrac{d^2y}{dx^2} + 4y = 0$ .

Auxiliary equation: $m^2 + 4 = 0$ So $m = \pm 2i$ .

Y = A\cos 2x + B\sin 2x

Example: Solve $\dfrac{d^2y}{dx^2} + 6\dfrac{dy}{dx} + 9y = 0$ .

Auxiliary equation: $m^2 + 6m + 9 = 0$ So $(m + 3)^2 = 0$ Giving $m = -3$ (repeated).

Y = (Ax + B)e^{-3x}

Non-Homogeneous Second Order Equations

A\frac{d^2y}{dx^2} + b\frac{dy}{dx} + cy = f(x)

General Solution: $y = y_h + y_p$ (complementary function + particular integral).

Method for finding $y_p$ : Guess the form of $y_p$ based on $f(x)$ .

$f(x)$	Guess for $y_p$
$e^{kx}$	$Ce^{kx}$ (if $k$ is not a root)
$kx + b$	$Ax + B$
$\cos kx$ or $\sin kx$	$A\cos kx + B\sin kx$

Example: Solve $\dfrac{d^2y}{dx^2} - 3\dfrac{dy}{dx} + 2y = 4e^{3x}$ .

Complementary function: $m^2 - 3m + 2 = 0$ So $m = 1, 2$ .

Y_h = Ae^x + Be^{2x}

For the particular integral, try $y_p = Ce^{3x}$ :

9Ce^{3x} - 9Ce^{3x} + 2Ce^{3x} = 4e^{3x}

2C = 4 \implies C = 2

Y_p = 2e^{3x}

General solution: $y = Ae^x + Be^{2x} + 2e^{3x}$ .

Maclaurin Series

The Maclaurin series expands a function as a power series about $x = 0$ :

F(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!}x^n

Standard Maclaurin Series

E^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots

\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots

\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots

\ln(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots, \quad |x| \leq 1, x \neq -1

(1 + x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots, \quad |x| < 1

Deriving Maclaurin Series

Example: Find the Maclaurin series for $f(x) = e^{-x^2}$ up to the term in $x^6$ .

Substitute $-x^2$ into the series for $e^u$ :

E^{-x^2} = 1 + (-x^2) + \frac{(-x^2)^2}{2!} + \frac{(-x^2)^3}{3!} + \cdots

= 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6} + \cdots

Applications of Maclaurin Series

Limits: Maclaurin series can be used to evaluate limits that are indeterminate.

Example: Evaluate $\displaystyle\lim_{x \to 0} \frac{e^x - 1 - x}{x^2}$ .

E^x - 1 - x = \left(1 + x + \frac{x^2}{2} + \cdots\right) - 1 - x = \frac{x^2}{2} + \cdots

\frac{e^x - 1 - x}{x^2} = \frac{\frac{x^2}{2} + \cdots}{x^2} = \frac{1}{2} + \cdots \to \frac{1}{2}

Integration of Series:

Example: Find $\displaystyle\int_0^{0.5} e^{-x^2} \, dx$ to 4 decimal places.

E^{-x^2} \approx 1 - x^2 + \frac{x^4}{2} - \frac{x^6}{6}

\int_0^{0.5} e^{-x^2} \, dx \approx \left[x - \frac{x^3}{3} + \frac{x^5}{10} - \frac{x^7}{42}\right]_0^{0.5}

= 0.5 - \frac{0.125}{3} + \frac{0.03125}{10} - \frac{0.0078125}{42}

= 0.5 - 0.04167 + 0.003125 - 0.000186 = 0.46127

Intuition

Calculus is the mathematics of change — differentiation finds the instantaneous rate of change (how fast something is happening right now), and integration finds the accumulation of change (how much has happened over time). The Fundamental Theorem of Calculus is the bridge between these two operations: they are inverses of each other, just as multiplication and division are inverses. Implicit differentiation is like working backwards from a recipe to find the ingredients: you know the result (the equation) and you work out how each variable contributes. The key insight is that calculus gives you exact answers about quantities that are constantly changing.

Worked Examples

See the examples integrated throughout the sections above.

Common Pitfalls

Forgetting the chain rule in implicit differentiation: When differentiating $y^3$ The result is $3y^2 \dfrac{dy}{dx}$ Not $3y^2$ .
Wrong choice of $u$ in integration by parts: Follow the LIATE rule. Choosing algebraic functions as $dv$ instead of $u$ leads to more complicated integrals.
Missing the constant of integration: Always include $+C$ for indefinite integrals.
Incorrect auxiliary equation: For $\dfrac{d^2y}{dx^2} + 4y = 0$ The auxiliary equation is $m^2 + 4 = 0$ Not $m^2 + 4m = 0$ .
Domain restrictions in Maclaurin series: The series for $\ln(1 + x)$ is valid for $-1 < x \le 1$ Not all $x$ .
Forgetting to change limits in definite substitution: When $u = g(x)$ The new limits are $u(a)$ and $u(b)$ Not $a$ and $b$ .
Not checking that the particular integral guess works: If your guess for $y_p$ contains a term already in $y_h$ Multiply by $x$ and try again.

Practice Questions

Find $\dfrac{dy}{dx}$ for $x^3 + xy^2 + y^3 = 7$ .
Evaluate $\displaystyle\int x^2 e^{-x} \, dx$ by parts.
Solve the differential equation $\dfrac{dy}{dx} = \dfrac{2x + 1}{3y - 2}$ with $y(0) = 1$ .
Find the Maclaurin series for $f(x) = \cos(2x)$ up to the term in $x^6$ .
Solve $\dfrac{d^2y}{dx^2} + 2\dfrac{dy}{dx} + y = 0$ .
A curve has parametric equations $x = t^2$$y = t^3 - 3t$ . Find the coordinates of the stationary points.
Evaluate $\displaystyle\int_0^1 \frac{dx}{1 + x^3}$ by first finding the partial fraction decomposition.
Solve $\dfrac{dy}{dx} + \dfrac{2y}{x} = x^3$ using an integrating factor.
Solve $\dfrac{d^2y}{dx^2} - 4y = 2e^{3x}$ .
Find the volume of revolution when $y = \sin x$ is rotated about the $x$ -axis from $x = 0$ to $x = \pi$ .
Evaluate $\displaystyle\int e^x \sin x \, dx$ using cyclic integration by parts.
Use Maclaurin series to evaluate $\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x^2}$ .

flowchart TD
    A[3_Calculus] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary

This topic covers the mathematical techniques and concepts related to further calculus, including key theorems, methods, and problem-solving approaches.

Key concepts include:

differentiation from first principles
product, quotient, and chain rules
integration techniques (by parts, substitution)
differential equations
applications to kinematics

Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.

Further Calculus | Highers - Wyatt's Notes

See Also