Further Calculus

This chapter covers Advanced Higher Mathematics content.

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

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 $\displaystyle\lim_{x \to 0} \frac{1 - \cos x}{x^2}$ .

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.

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