This video explains inverse trigonometric functions — arcsin, arccos, and arctan — and how they differ from regular trig functions like sin, cos, and tan.

Function vs. Inverse Function

A function $f$ takes an input and gives one output: $f(x) = y$. An inverse function $f^{-1}$ reverses that — it takes the output and gives back the original input: $f^{-1}(y) = x$.

Key idea: if $f$ and $f^{-1}$ undo each other, then $f^{-1}(f(x)) = x$.

Important condition: a function only has an inverse if it’s one-to-one (each output comes from exactly one input). This matters a lot for trig functions, since $\sin(x)$, $\cos(x)$, $\tan(x)$ repeat their values over and over — so we have to restrict their domain before we can invert them.

Notation

• $f^{-1}(x)$ means “the inverse function of $f$” — not $\frac{1}{f(x)}$
• Same for trig: $\sin^{-1}(x) \neq \frac{1}{\sin(x)}$. That’s why many textbooks prefer arcsin, arccos, arctan — it avoids the confusion entirely

Arcsine, Arccosine, Arctangent

In plain terms: $\arcsin(x)$ answers the question “what angle has a sine of $x$?” — same logic for the other two.

Example: $\sin(30°) = 0.5$, so $\arcsin(0.5) = 30°$.

This video walks through the basic differentiation rule, $\frac{d}{dx}(x^{n}) = nx^{n-1}$, and how it applies to polynomial functions.

Summary & Examples

• Derivative of $y=f(x)$ with respect to $x$ is denoted by $\frac{dy}{dx}, \frac{df}{dx}, f^{\prime}(x)$.

Power Rule

(a) $f(x)=x^{6}$
(b) $g(q)=\sqrt{q}$
(c) $h(p)=\frac{1}{p^{2}}$

Solution:

(a) \(\frac{df}{dx}=6x^{5}\)

(b) \(g(q)=\sqrt{q}=q^{1/2} \Rightarrow g^{\prime}(q)=\frac{1}{2}q^{-1/2}\)

(c) \(h(p)=\frac{1}{p^{2}}=p^{-2} \Rightarrow h^{\prime}=-2p^{-3}\)

Constant Multiple Rule

(a) $h(x)=5x^{3}$

(b) $g(x)=2\ln x$

An introduction to the Product Rule in differentiation — how to find the derivative of a function that is the product of two other functions.

Worked Example

Find the derivative of $f(x)=x^{2}\sin(x)$.

Solution:

The product rule states that for two differentiable functions $u(x)$ and $v(x)$:

Step 1 : Identify the components:

• $u(x)=x^{2} \Longrightarrow u^{\prime}(x)=2x$
• $v(x)=\sin(x) \Longrightarrow v^{\prime}(x)=\cos(x)$

Step 2 : Apply the product rule:

Step 3 : Simplify (optional):

The Quotient Rule tells you how to differentiate a function that is one function divided by another.

Formula

A common way to remember it: “low d-high minus high d-low, over the square of what’s below” — where “high” = $u$ (numerator) and “low” = $v$ (denominator).

How to Apply It

Step 1 — Identify the components: Split the function into $u(x)$ (top) and $v(x)$ (bottom), then find $u’(x)$ and $v’(x)$ separately.

Step 2 — Plug into the formula: \(f'(x) = \frac{u'v - uv'}{v^2}\)

Step 3 — Simplify (combine like terms in the numerator, factor if possible).

Worked Example

Find the derivative of $f(x) = \dfrac{\sin(x)}{x}$.

• $u(x) = \sin(x) \Rightarrow u’(x) = \cos(x)$
• $v(x) = x \Rightarrow v’(x) = 1$

Common mistake to watch for: the order in the numerator matters — it’s $u’v - uv’$, not $uv’ - u’v$. Swapping the order flips the sign of the whole answer.

Integration is the reverse of differentiation. For each standard function, remember its antiderivative plus a constant $C$ (since the derivative of any constant is 0).

Standard Functions

Key points to note:

• $x^k$ rule breaks when $k = -1$ (that’s exactly when $1/x$ takes over, giving $\ln

  x

  $ instead)

• $e^x$ is unique — it integrates to itself
• Sign flips for $\sin x$: integrating gives $-\cos x$, not $+\cos x$ (easy to mix up)
• Always add $+C$ — dropping it is the most common mistake

Example: $\int t^3\, dt = \dfrac{t^4}{4} + C$

When a Function Doesn’t Match a Standard Form

If the integral isn’t directly one of these standard results, a technique is needed to reduce it to one. Three common methods (not an exhaustive list):

1. Integration by Substitution — rewrite the integral in terms of a new variable to simplify it into a standard form
2. Integration by Parts — used for a product of two functions
3. Partial Fractions — break a complex rational function into simpler fractions, then integrate each term using the standard results above

This method works when the integrand contains a function $f(x)$ and its derivative $f’(x)$ together. Once you recognize the pattern, you can integrate directly — no formal substitution needed.

Three Standard Forms

1. \(\int \frac{f'(x)}{f(x)}\,dx = \ln\lvert f(x) \rvert + C\)

Derivation: since $\dfrac{d}{dx}\ln\lvert f \rvert = \dfrac{f’}{f}$, integrating both sides gives the result.

2. \(\int [f(x)]^n f'(x)\,dx = \frac{[f(x)]^{n+1}}{n+1} + C\)

Derivation: let $u = f(x)$, so $du = f’(x)\,dx$. This turns the integral into $\int u^n\,du = \dfrac{u^{n+1}}{n+1} + C$.

3. \(\int e^{f(x)} f'(x)\,dx = e^{f(x)} + C\)

Derivation: since $\dfrac{d}{dx}e^f = e^f \cdot f’$, integrating both sides gives the result.

Worked Example 1 — Recognition Method

Step 1 — Identify $f(x)$ and $f’(x)$: Let $f(x) = e^{2x} - 2$, so $f’(x) = 2e^{2x}$.

Step 2 — Rewrite the numerator to match $f’(x)$: \(e^{2x} = \frac{1}{2}f'(x)\)

Step 3 — Apply Formula 1: \(\int \frac{e^{2x}}{e^{2x}-2}\,dx = \frac{1}{2}\int \frac{f'(x)}{f(x)}\,dx = \frac{1}{2}\ln\lvert e^{2x}-2 \rvert + C\)

Key point: the constant $\frac{1}{2}$ is pulled out to make the numerator match $f’(x)$ exactly — this adjustment step is the main skill to practice with the recognition method.

Worked Example 2 — Change of Variable

Not every integral fits the recognition shortcut directly. Sometimes it’s clearer to formally substitute a new variable $u$, rewrite the entire integral in terms of $u$ (including $x$ itself), integrate, then substitute back.

Step 1 — Choose the substitution: Let $u = x+1$, so $x = u-1$ and $du = dx$.

Step 2 — Rewrite the whole integral in terms of $u$: \(\int x\sqrt{x+1}\,dx = \int (u-1)\sqrt{u}\,du = \int \left(u^{3/2} - u^{1/2}\right) du\)

Step 3 — Integrate using the Power Rule: \(= \frac{2}{5}u^{5/2} - \frac{2}{3}u^{3/2} + C\)

Step 4 — Substitute back $u = x+1$: \(\int x\sqrt{x+1}\,dx = \frac{2}{5}(x+1)^{5/2} - \frac{2}{3}(x+1)^{3/2} + C\)

Key point: unlike Example 1, here $x$ itself had to be rewritten in terms of $u$ (as $u-1$) before integrating — this is the defining feature of the change-of-variable method, as opposed to just spotting a ready-made $f$ and $f’$ pair.

This video explains Integration by Parts — a technique for integrating a product of two functions when standard formulas or substitution don’t directly work.

Formula

Used when integrating a product of two functions that doesn’t fit substitution.

How to Choose u and dv

A common guide is LIATE — pick $u$ in this priority order: Logarithmic, Inverse trig, Algebraic, Trigonometric, Exponential. Whichever type appears first in that list becomes $u$; the rest becomes $dv$.

Worked Example

Find $\int x\,e^x\,dx$.

Step 1 — Choose $u$ and $dv$: By LIATE, $x$ (algebraic) ranks above $e^x$ (exponential), so: \(u = x, \quad dv = e^x\,dx\)

Step 2 — Find $du$ and $v$: \(du = dx, \quad v = e^x\)

Step 3 — Apply the formula: \(\int x\,e^x\,dx = x\,e^x - \int e^x\,dx\)

Step 4 — Integrate what’s left and simplify: \(\int x\,e^x\,dx = x\,e^x - e^x + C\)

Key point: the whole trick is choosing $u$ so that $\int v\,du$ ends up simpler than the original integral — here, differentiating $x$ down to $1$ (via $du$) is what made the leftover integral trivial.

A beginner’s walkthrough of the Maclaurin series — how to build a polynomial approximation of a function purely from its derivatives at zero.

The Idea

A Maclaurin series rewrites a function as an infinite polynomial, built entirely from the function’s derivatives at $x = 0$.

It’s a special case of the Taylor series, centred at $0$ instead of some general point $a$.

Worked Example: Deriving $e^x$

Take $f(x) = e^x$. Every derivative of $e^x$ is $e^x$, so:

Substituting into the formula:

Worked Example: Deriving $\sin x$

Take $f(x) = \sin x$. Differentiate repeatedly and evaluate at $x=0$:

The pattern repeats every 4 derivatives ($\sin \to \cos \to -\sin \to -\cos \to \sin \cdots$), and every even-order term is $0$.

Only odd powers survive — a direct consequence of $\sin x$ being an odd function.

Standard Expansions Table

Pattern to notice:

• $\sin x$ and $\cos x$ only ever have odd or even powers respectively — because $\sin$ is an odd function and $\cos$ is an even function.
• $\ln(1-x)$ is just $\ln(1+x)$ with every $x \to -x$ — same trick works for $\dfrac{1}{1-x}$ vs $\dfrac{1}{1+x}$.
• $\dfrac{1}{1-x}$ is the geometric series you already know: $1 + x + x^2 + \cdots$

Definition

A function $f(x, y)$ is homogeneous of degree $n$ if, for every $t$,

Quick test (no need to substitute $t$): add up the powers of $x$ and $y$ in each term. If every term has the same total degree, the function is homogeneous of that degree.

Euler’s Theorem

If $f(x,y)$ is homogeneous of degree $n$, then

where $f_x = \dfrac{\partial f}{\partial x}$ and $f_y = \dfrac{\partial f}{\partial y}$.

In words: scale each partial derivative by the variable it came from, add them up, and the result is just $n$ times the original function. This gives a quick way to check homogeneity once the derivatives are known, without repeating the $f(tx,ty) = t^n f(x,y)$ substitution.

Worked Example

Question: Show that

is homogeneous, and verify that Euler’s equation holds.

Solution

Step 1 — Check homogeneity. Replace $x \to tx$, $y \to ty$:

Simplify each term:

• $(tx)^3\sin\left(\dfrac{tx}{ty}\right) = t^3x^3\sin\left(\dfrac{x}{y}\right)$ — the $t$’s cancel inside the sine
• $(tx)^2\left[t^2(x^2+y^2)\right]^{1/2} = t^2x^2 \cdot t(x^2+y^2)^{1/2} = t^3x^2(x^2+y^2)^{1/2}$

So

$f$ is homogeneous of degree $n = 3$.

Step 2 — Compute the partial derivatives.

Step 3 — Form $x f_x + y f_y$.

Multiply $f_x$ by $x$:

Multiply $f_y$ by $y$:

Add them — the cosine terms cancel exactly:

Step 4 — Simplify the remaining term. Factor the numerator, $x^4 + x^2y^2 = x^2(x^2+y^2)$:

Substituting back:

Conclusion

Euler’s theorem is verified, consistent with $f$ being homogeneous of degree $n = 3$.

What is the Lagrange Multiplier Method?

The Lagrange multiplier method finds the maximum or minimum of a function subject to a constraint — a condition the variables must satisfy.

• The objective function is the quantity you’re trying to maximize or minimize, e.g. $f(x,y)$ or $f(x,y,z)$
• The constraint is an equation relating the variables that must hold, usually written as $g(x,y)=k$ or $g(x,y,z)=k$

How to identify which is which: the objective is what the question asks you to maximize/minimize; the constraint is the “where…” or “subject to…” condition — often a fixed total, area, volume, or sum.

The idea: at an extremum on the constraint surface, the gradient of $f$ is parallel to the gradient of $g$. This gives the condition $\nabla f = \lambda \nabla g$, where $\lambda$ (lambda) is the Lagrange multiplier.

For two variables, maximize/minimize $f(x,y)$ subject to $g(x,y)=k$:

Solve these three equations simultaneously for $x, y, \lambda$.

For three variables, maximize/minimize $f(x,y,z)$ subject to $g(x,y,z)=k$:

Solve these four equations simultaneously for $x, y, z, \lambda$.

Worked Example

Question

Apply the Lagrange multiplier method to find $x$, $y$ and $z$ to maximize

where $x$, $y$ and $z$ are positive numbers, and the sum is equal to $6$.

Solution

Step 1 — Simplify with logs

Since $f>0$, maximizing $f$ is the same as maximizing $\ln f$ (a monotonic transform), which turns the product into a sum — much easier to differentiate:

Constraint: $g(x,y,z) = x+y+z-6=0$

Step 2 — Set up the Lagrangian

Step 3 — Take partial derivatives and set to zero

By symmetry, the same form holds for $y$ and $z$. The Lagrange conditions give:

Step 4 — Argue x = y = z

The function $h(t) = \sqrt{t}(1+\sqrt{t})$ is strictly increasing for $t>0$ (both $\sqrt t$ and $1+\sqrt t$ increase). Since $h(x)=h(y)=h(z)$, this forces:

Step 5 — Apply the constraint

Step 6 — Compute the maximum value (surd form)

Expand using $(1+\sqrt2)^2 = 3+2\sqrt2$, then $(1+\sqrt2)^3 = (1+\sqrt2)(3+2\sqrt2) = 7+5\sqrt2$, so:

Step 7 — Confirm it’s a maximum

Checking a boundary case such as $x\to0,\ y=z=3$ gives $f=(1+\sqrt3)^4$, which is a much smaller value — confirming the symmetric interior point is the maximum.

Answer: $x=y=z=2$, giving a maximum value of $f_{\max} = 99+70\sqrt2$.

Matrix Form and Augmented Matrix

For a system:

The augmented matrix (last column = constants):

The Three Row Operations

1. Multiply a row by a non-zero constant
2. Add or subtract one row with another
3. Exchange two rows

Worked Example

Back-substitution gives $z=3$, $y=2$, $x=1$.

This video covers how to solve a system of three linear equations in three unknowns using matrix methods, and how to tell — before fully solving — whether the system has a unique solution, infinitely many solutions, or no solution at all.

Setting Up the Matrix Form

A system of three equations rewrites into matrix form $A\mathbf{x} = \mathbf{b}$:

For row operations, it’s more convenient to work with the augmented matrix — $A$ and $\mathbf{b}$ side by side, with $\mathbf{b}$ as the final column:

Which of the three cases below you land in depends entirely on what happens to this matrix.

Case 1 — Unique Solution

Test: compute the determinant of $A$. If $\det(A) \neq 0$, the system has exactly one solution.

Determinant formula (3×3):

Geometric picture: each equation represents a plane in 3D space. Three planes with $\det(A) \neq 0$ intersect at exactly one point — that point is the unique solution.

Worked example — same system solved three ways:

Method A — Cramer’s Rule

Replace one column of $A$ with $\mathbf{b}$ at a time, and divide by $\det(A)$:

Method B — Inverse Matrix Method

For this $A$, the adjugate works out to:

Method C — Row Operations (Gaussian Elimination)

Back-substitute: $z=3$, then $-3y-3=-9 \Rightarrow y=2$, then $x+2+3=6 \Rightarrow x=1$.

All three methods agree: $(x,y,z) = (1,2,3)$ — the single point where the three planes meet.

Case 2 — Infinitely Many Solutions (Consistent, Dependent)

Test: $\det(A) = 0$, and row reduction produces a full row of zeros, including the right-hand side — one equation is redundant, carrying no new information.

Geometric picture: the three planes intersect along a common line rather than a single point — every point on that line satisfies all three equations.

Worked example:

The full zero row confirms infinitely many solutions. Let $z = k$ (free parameter):

General solution: $\left(\dfrac{11-3k}{2},\ \dfrac{k+1}{2},\ k\right)$ for any real $k$. Check $k=1$: $(4,1,1)$ satisfies all three original equations.

Case 3 — No Solution (Inconsistent)

Test: $\det(A) = 0$, but row reduction produces a row where the left side is all zeros while the right-hand side is non-zero — a contradiction like $0 = 1$.

Geometric picture: the three planes don’t share a common point or line — for example, two planes may meet along a line that runs parallel to the third plane, never touching it.

Worked example — same $A$ as Case 2, only the last equation’s constant changes:

The last row reads $0x+0y+0z = 1$ — impossible. No solution exists.

Quick Reference

First Order ODE — Integrating Factor Method

General form:

Integrating factor:

Procedure

1. Write the equation in standard form $y’ + P(x)y = Q(x)$
2. Compute $\mu = e^{\int P\,dx}$
3. Multiply through by $\mu$ — left side becomes $\dfrac{d}{dx}(\mu y)$
4. Integrate both sides
5. Divide by $\mu$ to isolate $y$
6. Apply an initial condition (if given) to find $C$

Worked Example

Applying $y(0)=3$ gives $C=2$:

Second Order ODE — Constant Coefficients (Homogeneous)

General form:

Auxiliary equation:

The discriminant $b^2-4ac$ determines the case:

Case 1 — Distinct real roots ($b^2-4ac>0$): $y = C_1e^{r_1x} + C_2e^{r_2x}$

Example: $y’’ - y’ - 6y = 0 \;\Rightarrow\; r=3,-2$

Solution:

Case 2 — Repeated root ($b^2-4ac=0$): $y = (C_1 + C_2x)e^{rx}$

Example: $y’’ - 6y’ + 9y = 0 \;\Rightarrow\; r=3$ (double)

Solution:

Case 3 — Complex roots ($b^2-4ac<0$), $r=\alpha\pm i\beta$: $y = e^{\alpha x}(C_1\cos\beta x + C_2\sin\beta x)$

Example: $y’’ + 4y’ + 13y = 0 \;\Rightarrow\; r=-2\pm 3i$

Solution:

General Form and Solution Structure

• $y_c$ — complementary solution (solve the homogeneous version)
• $y_p$ — particular solution (trial function matching the shape of $f(x)$)

Standard trial forms

Resonance rule: if the trial already appears in $y_c$, multiply it by $x$ (simple root) or $x^2$ (repeated root).

Example 1 — No Overlap

Solve the differential equation

given the initial conditions $y(0)=2$ and $y’(0)=3$.

Solution:

Example 2 — Overlap, Simple Root

Solve the differential equation

given the initial conditions $y(0)=1$ and $y’(0)=0$.

Solution:

Example 3 — Overlap, Repeated Root

Solve the differential equation

given the initial conditions $y(0)=1$ and $y’(0)=2$.

Solution:

This example shows how diagonalizing a matrix — finding $P$ and $D$ such that $P^{-1}AP = D$ — can be used to decouple and solve a system of linear differential equations.

Example

Find the eigenvalues of the matrix \(A = \begin{pmatrix} 11 & -6 \\ 18 & -10 \end{pmatrix}\), and find an eigenvector corresponding to each eigenvalue. Hence find an invertible matrix $P$ and a diagonal matrix $D$ such that $P^{-1}AP = D$.

Use your result to find the functions $f(t)$ and $g(t)$ that satisfy the differential equations

with the initial conditions $f(0)=2$ and $g(0)=1$.

Solution:

Step 1 — Find the Eigenvalues

Solve $\det(A - \lambda I) = 0$:

Step 2 — Eigenvector for $\lambda_1 = 2$

Pick $x=2$:

Step 3 — Eigenvector for $\lambda_2 = -1$

Pick $x=1$:

Step 4 — Build $P$ and $D$

$\det(P) = 1$, so:

Step 5 — Decouple the Equations

With $\mathbf{x} = \begin{pmatrix} f \ g \end{pmatrix}$ and $\mathbf{x} = P\mathbf{y}$, since $P^{-1}AP=D$, the coupled system becomes two independent equations:

Step 6 — Apply Initial Conditions

Step 7 — Convert Back to $f(t)$ and $g(t)$

Since $\mathbf{x}=P\mathbf{y}$: $f = 2u+v$, $g = 3u+2v$. Substituting $u=3e^{2t}$, $v=-4e^{-t}$:

Step 8 — Verify

• $f(0)=6-4=2$ ✓, $g(0)=9-8=1$ ✓
• $f’(t)=12e^{2t}+4e^{-t} = 11f-6g$ ✓
• $g’(t)=18e^{2t}+8e^{-t} = 18f-10g$ ✓

Learning Objective

Learn how to solve a second-order linear difference equation when the characteristic equation has complex roots. You’ll find the modulus and argument (basic angle) of the roots using your calculator, then build the closed-form solution using cosine and sine terms.

The Method

For a recurrence $a_{n+2} + pa_{n+1} + qa_n = 0$ with characteristic equation $x^2+px+q=0$: if the discriminant is negative, the roots are $\alpha \pm i\beta$, and

where $r = \sqrt{\alpha^2+\beta^2}$ and $\theta = \cos^{-1}\left(\dfrac{\alpha}{r}\right)$ (adjusted for sign of $\beta$).

Worked Example

Question

A sequence satisfies the recurrence

Find a closed-form expression for $a_n$.

Solution

Step 1 — Characteristic equation

Step 2 — Solve with the quadratic formula

Complex conjugate roots: $\alpha = 1$, $\beta = 1$.

Step 3 — Modulus (calculator: √)

Step 4 — Basic angle (calculator: inverse cos)

($\beta>0$, so no sign adjustment needed.)

Step 5 — General solution

Step 6 — Apply $a_0=1$

Step 7 — Apply $a_1=1$

Step 8 — Final closed form

Step 9 — Verify

$a_2 = 2\cos(\pi/2)=0$, matches $2a_1-2a_0=0$ ✓ $a_3 = 2\sqrt2\cos(3\pi/4)=-2$, matches $2a_2-2a_1=-2$ ✓

General form

Example

Question: Solve the difference equation

and find $y_n$ in terms of $n$.

Step 1 – Identify a and b. The form is $y_{n+1} = a\,y_n + b$, so $a = 2$ and $b = 3$.

Step 2 – Find the equilibrium (the value that never changes). Set $y_{n+1} = y_n = y^*$:

Step 3 – Solve the homogeneous part. Drop the constant: $y_{n+1} = 2y_n$. This gives

Step 4 – Write the general solution.

Step 5 – Use the starting value to find A. Put $n = 0$ and $y_0 = 1$:

Step 6 – Final answer.

Step 7 – Check by direct calculation.

Stability

Here $\lvert a \rvert = 2 > 1$, so the sequence grows without limit and moves away from $y^* = -3$.

If $\lvert a \rvert < 1$, the sequence settles at the equilibrium instead.

Example: A sequence of numbers $x_1, x_2, x_3 \ldots$ is given by

Find an explicit formula for $x_t$:

Method 1: Apply Formula

Auxiliary equation: $m^{2}+3m+3=0$ where $a=1, b=3, c=3$

Complementary function:

Method 2: Geometric Approach

Auxiliary Equation: $m^{2}+3m+3=0$ where $a=1, b=3, c=3$

Complementary function $\Rightarrow x_{t}=(\sqrt{3})^{t}\left[A\cos\left(\frac{5\pi}{6}t\right)+B\sin\left(\frac{5\pi}{6}t\right)\right]$