Racquel Sanderson

February 3, 2025 · 9:11 am

Differentiating Trigonometric Functions

In the last post we looked at two trig limits:

(1) $\begin{equation*}\lim_{x \to 0}\frac{sin(x)}{x}=1\end{equation*}$

(2) $\begin{equation*}\lim_{x \to 0}\frac{1-cos(x)}{x}=0\end{equation*}$

We are going to use these two limits to differentiate sine and cosine functions from first principals.

$\begin{equation*}f(x)=sin(x)\end{equation}$

$\begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{sin(x+h)-sin(x)}{h}\end{equation}$

Use the trig identity

$\begin{equation*}sin(A+B)=sinAcosB+sinBcosA\end{equation}$

$\begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{sin(x)cos(h)+sin(h)cos(x)-sin(x)}{h}\end{equation}$

$\begin{equation*}f'(x)=\lim\limits_{h \to 0}(\frac{sin(x)(cos(h)-1)}{h}+\frac{sin(h)cos(x)}{h})\end{equation}$

$\begin{equation*}f'(x)=sin(x)\lim\limits_{h \to 0}(\frac{(cos(h)-1)}{h}+cos(x)\lim\limits_{h \to 0}\frac{sin(h)}{h}\end{equation}$

$\begin{equation*}f'(x)=sin(x)\lim\limits_{h \to 0}(\frac{-(-cos(h)+1)}{h}+cos(x)\lim\limits_{h \to 0}\frac{sin(h)}{h}\end{equation}$

Evaluate the limits

$\begin{equation*}f'(x)=sin(x)\times 0+cos(x)\times (1)=cos(x)\end{equation}$

Hence, $\frac{d}{dx}sin(x)=cos(x)$ .

Now we are going to do the same for $f(x)=cos(x)$ .

$\begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{cos(x+h)-cos(x)}{h}\end{equation}$

Use the trigonometric identity

$\begin{equation*}cos(A+B)=cosAcosB-sinAsinB\end{equation}$

$\begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{cos(x)cos(h)-sin(x)sin(h)-cos(x)}{h}\end{equation}$

$\begin{equation*}f'(x)=\lim\limits_{h \to 0}\frac{cos(x)(cos(h)-1)-sin(x)sin(h)}{h}\end{equation}$

$\begin{equation*}f'(x)=cos(x)\lim\limits_{h \to 0}\frac{-(1-cos(h))}{h}-sin(x)\lim\limits_{h \to 0}\frac{sin(h)}{h}\end{equation}$

Evaluate the limits

$\begin{equation*}f'(x)=cos(x)\times(0)-sin(x)\times (1)=-sin(x)\end{equation}$

Hence $\frac{d}{dx} cos(x)=-sin(x)$

1 Comment

Filed under Calculus, Differentiation, Identities, Trigonometry, Year 12 Mathematical Methods

Tagged as differentiation, trig differentiation, trig identities, trig limits, trigonometric differentiation, trigonometric identities, trigonometric limits

February 2, 2025 · 7:46 am

Trigonometric Limits

$\lim\limits_{x \to 0}\frac{sin(x)}{x}=?$

Remember $cos(x)=\frac{OA}{OB}=\frac{OA}{1}$ , hence $OA=cos(x)$ and the co-ordinate of $A$ is $(cos(x), 0)$ .

$sin(x)=\frac{AB}{OB}=\frac{AB}{1}$ , hence $AB=sin(x)$ and the co-ordinate of $B$ is $(cos(x), sin(x))$

And from the definition of $tan(x)$ we know $D$ is the point $(1, tan(x))$

Consider the areas of triangle $OAB$ , sector $OBC$ , and triangle $OCD$ .

We know from inspection of the above diagram that

Area $OAB<$ Area $OCB<$ Area $OCD$

Which means,

$\frac{1}{2}b_1 h_1<\frac{1}{2}r^2x<\frac{1}{2}b_2 h_2$

We can ignore all of the halves.

$cos(x)sin(x)<x<(1)tan(x)$

Remember $tan(x)=\frac{sin(x)}{cos(x)}$

$cos(x)sin(x)<x<\frac{sin(x)}{cos(x)}$

Divide everything by $sin(x)$ (as we are in the first quadrant we know $sin(x)>0$ , so we don’t need to worry about the inequality)

$cos(x)<\frac{x}{sin(x)}<\frac{1}{cos(x)}$

Invert everything and change the direction of the inequalities)

$\frac{1}{cos(x)}>\frac{sin(x)}{x}>cos(x)$

I am going to rewrite it as follows

$cos(x)<\frac{sin(x)}{x}<\frac{1}{cos(x)}$

because I like to use less thans rather than greater thans.

Now what happens as $x$ tends to $0$ ?

$cos(0)=1$

$1<\frac{sin(x)}{x}<\frac{1}{1}$

Hence by the squeeze theorem $\lim\limits_{x \to 0}\frac{sin(x)}{x}=1$

Now we know this limit, we are going to use it to find $\lim\limits_{x \to 0}\frac{1-cos(x)}{x}$

Multiply by $\frac{1+cos(x)}{1+cos(x)}$

$\lim\limits_{x \to 0}\frac{1-cos(x)}{x}\times \frac{1+cos(x)}{1+cos(x)}$

$\lim\limits_{x \to 0}\frac{1-cos^2(x)}{x(cos(x)+1)}$

$\lim\limits_{x \to 0}\frac{sin^2(x)}{x(cos(x)+1)}$

$\lim\limits_{x \to 0}\frac{sin(x)}{x}\times sin(x)(cos(x)+1)}$

$\lim\limits_{x \to 0}\frac{sin(x)}{x}\times \lim\limits_{x \to 0}sin(x)(cos(x)+1)$

If we evaluate the limits,

$(1)(sin(0)(cos(0)+1)=1\times 0 \times 2=0$

Hence, $\lim\limits_{x \to 0}\frac{1-cos(x)}{x}=0$

In the next post we are going to use these limits to differentiate sine and cosine functions.

1 Comment

Filed under Area, Area of Triangles (Sine), Calculus, Identities, Trigonometry, Year 12 Mathematical Methods

Tagged as trig limits, trigonometric limits

January 16, 2025 · 8:08 am

Two circles have the same centre $O$ . The smaller circle has a radius of $1$ cm, while the larger has a radius of $(x+1)$ cm. The circles enclose a region $QRST$ , which is subtended by angle of ${\theta}$ at $O$ , as shaded.

The area of $QRST$ is $A$ cm², where $A$ is a constant and $A>0$

Let $P$ cm be the perimeter of $QRST$

(a) By finding expressions for the area and perimeter of $QRST$ show that $P(x)=2x+\frac{2A}{x}$

(b) Show that if the perimeter is minimised, then ${\theta}$ must be less than $2$ .

Leave a Comment

Filed under Differentiation, Optimisation, Uncategorized

Tagged as differentiation, HSC Advanced 2024, optimising

January 14, 2025 · 7:52 am

Deriving the Quotient Rule for Differentiation

Like we did for the product rule, we are going to derive the differentiating rule for functions in the form $y=\frac{f(x)}{g(x)}$ .

Something like, $y=\frac{x^2+3x+2}{x^3-1}$

Remember the first principals limit

$\lim_{\limits h \to 0}\frac{f(x+h)-f(x)}{h}$

If $y=\frac{f(x)}{g(x)}$ , then

$y'=\lim_{\limits h \to 0}\frac{\frac{f(x+h)}{g(x+h)}-\frac{f(x)}{g(x)}}{h}$

Find a common denominator for the numerator (i.e. $g(x+h)g(x)$ )

$y'=\lim_{\limits h \to 0}\frac{\frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)}}{h}$

To make things a bit easier I am going to multiply by $\frac{1}{h}$ rather than having $h$ as the denominator

$y'=\lim_{\limits h \to 0}\frac{f(x+h)g(x)-f(x)g(x+h)}{g(x+h)g(x)} \times \frac{1}{h}$

Now I am going to add and subtract $f(x)g(x)$

$y'=\lim_{\limits h \to 0}\frac{f(x+h)g(x)-f(x)(g(x)+f(x)g(x)-f(x)g(x+h)}{g(x+h)g(x)} \times \frac{1}{h}$

Factorise

$y'=\lim_{\limits h \to 0}\frac{g(x)(f(x+h)-f(x))+f(x)(g(x)-g(x+h))}{g(x+h)g(x)} \times \frac{1}{h}$

Change the sign in the middle

$y'=\lim_{\limits h \to 0}\frac{g(x)(f(x+h)-f(x))-f(x)(g(x+h)-g(x))}{g(x+h)g(x)} \times \frac{1}{h}$

Separate the limits

$y'=g(x)\lim_{\limits h \to 0}\frac{\frac{f(x+h)-f(x)}{h}}{g(x+h)g(x)}-f(x)\lim_{\limits h \to 0}\frac{\frac{g(x+h)-g(x)}{h}}{g(x+h)g(x)}$

which simplifies to

$y'=g(x)\frac{f'(x)}{g(x)g(x)}-f(x)\frac{g'(x)}{g(x)g(x)}$

$y'=\frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}$

In words

The derivative of the top times the bottom take the derivative of the bottom times the top all over the bottom squared

Example

$y=\frac{x^2+3x+2}{x^3-1}$

$y'=\frac{(2x+3)(x^3-1)-3x^2(x^2+3x+2)}{(x^3-1)^2}$

$y'\frac{2x^4-2x+3x^3-3-3x^4-9x^3-6x^2}{(x^3-1)^2}$

$y'=\frac{-x^4-6x^3-6x^2-2x-3}{(x^3-1)^2}$

Exam questions usually specify no simplifying.

2 Comments

Filed under Calculus, Differentiation, Quotient Rule, Year 12 Mathematical Methods

January 11, 2025 · 9:56 am

Deriving the Product Rule for Differentiation

In my previous post we looked at the Chain Rule for Differentiation, this post is on the Product Rule. Differentiating a function in the form $y=f(x)\times g(x)$ .

For example, $y=(3x^3+2x-1)(x^4+2x^2)$

Remember differentiating from first prinicpals:

$f'(x)=\lim_{\limits h \to 0} \frac{f(x+h)-f(x)}{h}$

$y=f(x)g(x)$

$\frac{dy}{dx}=\lim_{\limits h\to 0}\frac{f(x+h)g(x+h)-f(x)g(x)}{h}}$

$\small{ \frac{dy}{dx}=\lim_{\limits h \to 0} \frac{f(x+h)g(x+h)-g(x+h)f(x)+g(x+h)f(x)-f(x)g(x)}{h}}$

By subtracting and then adding $g(x+h)f(x)$ we haven’t changed the limit, but it means we can do some factorising.

$\frac{dy}{dx}=\lim_{\limits h \to 0}\frac{g(x+h)(f(x+h)-f(x))+f(x)(g(x+h)-g(x))}{h}$

$\small{\frac{dy}{dx}=\lim_{\limits h \to 0}g(x+h)\lim_{\limits h \to 0}\frac{f(x+h)-f(x)}{h}+\lim_{\limits h \to 0}f(x)\lim_{\limits h \to 0}\frac{g(x+h)-g(x)}{h}}$

When we evaluate the limits

$\frac{dy}{dx}=g(x)f'(x)+f(x)g'(x)$

Example

Find the derivative of $y=(3x^3+2x-1)(x^4+2x^2)$

I remember the rule in words ‘derivative of the first times the second plus the derivative of the second times the first’.

$y'=(9x^2+2)(x^4+2x^2)+(4x^3+4x)(3x^3+2x-1)$

$y'=9x^6+18x^4+2x^4+4x^2+12x^6+8x^4-4x^3+12x^4+8x^2-4x$

$y'=21x^6+40x^4-4x^312x^2-4x$

Most exam questions have ‘don’t simplify’, so the first line of working above would be enough.

Onto the Quotient Rule.

2 Comments

Filed under Calculus, Differentiation, Product Rule, Year 12 Mathematical Methods

January 9, 2025 · 9:40 am

Deriving the Chain Rule for Differentiation

How to differentiate something in the form $y=[f(x)]^n$

For example, $y=(3x^2-2x+6)^5$ , we could expand the expression, but the Chain Rule provides a quick and easy method.

Differentiate $y=[f(x)]^n$

Let $u=f(x)$ , then $y=u^n$

We want to find $\frac{dy}{dx}$ , but $\frac{dy}{dx}=\frac{dy}{du}\times \frac{du}{dx}$

They’re not fractions, but limits of fractions, but they work like fractions.

$\frac{du}{dx}=f'(x)$ and $\frac{dy}{du}=nu^{n-1}$

Therefore, $\frac{dy}{dx}=f'(x)\times nu^{n-1}$

Replace $u$ with $f(x)$

(1) $\begin{equation*}\frac{dy}{dx}=n[f(x)]^{n-1}f'(x)\end{equation*}$

What about a function in the form $y=f(g(x))$ ?

We’re going to follow the same process.

Let $u=g(x)$ , then $y=f(u)$

$\frac{du}{dx}=g'(x)$ and $\frac{dy}{du}=f'(u)$

Therefore $\frac{dy}{dx}=f'(u)g'(x)$

(2) $\begin{equation*}\frac{dy}{dx}=f'(g(x))g'(x) \end{equation*}$

Equations $1$ and $2$ are versions of the Chain Rule.

Example

Find the derivative of $y=(3x^2-2x+6)^5$

$\begin{equation*}\frac{dy}{dx}=5(3x^2-2x+6)^4\times (6x-2)\end{equation}$

$\begin{equation*}\frac{dy}{dx}=5(6x-2)(3x^2-2x+6)^4\end{equation}$

$\begin{equation*}\frac{dy}{dx}=10(3x-1)(3x^2-2x+6)\end{equation}$

Next time we are going to look at the Product Rule.

1 Comment

Filed under Calculus, Chain Rule, Differentiation, Year 12 Mathematical Methods

Tagged as Calculus, chain rule, differentiation, year 12 mathematical methods

January 6, 2025 · 3:58 am

An Imaginary Tale – The story of i

An Imaginary Tale – The Story of $\sqrt{-1}$

I bought a second hand copy of this book from Abe Books (it’s possible to find cheap maths books this way).

Here’s the blurb …

Today complex numbers have such widespread practical use–from electrical engineering to aeronautics–that few people would expect the story behind their derivation to be filled with adventure and enigma. In An Imaginary Tale , Paul Nahin tells the 2000-year-old history of one of mathematics’ most elusive numbers, the square root of minus one, also known as i . He recreates the baffling mathematical problems that conjured it up, and the colorful characters who tried to solve them. In 1878, when two brothers stole a mathematical papyrus from the ancient Egyptian burial site in the Valley of Kings, they led scholars to the earliest known occurrence of the square root of a negative number. The papyrus offered a specific numerical example of how to calculate the volume of a truncated square pyramid, which implied the need for i . In the first century, the mathematician-engineer Heron of Alexandria encountered I in a separate project, but fudged the arithmetic; medieval mathematicians stumbled upon the concept while grappling with the meaning of negative numbers, but dismissed their square roots as nonsense. By the time of Descartes, a theoretical use for these elusive square roots–now called “imaginary numbers”–was suspected, but efforts to solve them led to intense, bitter debates. The notorious i finally won acceptance and was put to use in complex analysis and theoretical physics in Napoleonic times. Addressing readers with both a general and scholarly interest in mathematics, Nahin weaves into this narrative entertaining historical facts and mathematical discussions, including the application of complex numbers and functions to important problems, such as Kepler’s laws of planetary motion and ac electrical circuits. This book can be read as an engaging history, almost a biography, of one of the most evasive and pervasive “numbers” in all of mathematics.

It took me a long time to read this book – according to storygraph I started it in August (12th to be exact). I really enjoyed this book, but I do think you need to understand maths. I enjoyed working through all of the different formulae and examples. I have put some on this blog

I found the historical aspects very interesting.

Minus times minus is plus

The reason for this we need not discuss

W. H. Auden

Maybe I should use the above quote with my Year 8s who are just starting on their negative number journey.

This book covers quite complex (pun intended) ideas – particularly in Chapter 6 Wizard Mathematics and Chapter 7 The Nineteenth Century, Cauchy, and the Beginning of Complex Function Theory.

If you can do algebra and a bit of calculus and complex numbers interest you, then I think this book is for you.

A review

Leave a Comment

Filed under Book Review

January 1, 2025 · 9:40 am

Volume of Rotation About a Slanting Line

Given the area in the first quadrant bounded by $x^2=12y$ , the line $y=3$ and the $y-$ axis. What is the volume generated when this area is rotated about the line $2x-y+4=0$ ?

Rotate the green region about the line $y=2x+4$

Rotate the green region about the line $y=2x+4$

We can split the solid into shells.

$\begin{equation*}V=2\pi r dx dy\end{equation}$

Where $r$ is the distance from each $(x,y)$ point in the region to the line $2x-y+4=0$ , $dx$ is the width, and $dy$ is the height.

The distance between a point and a line is

$\begin{equation*} d=\frac{Ax+By+C=0}{\sqrt{A^2+B^2}}\end{equation}$

Hence, $r=\frac{2x-y+4}{\sqrt{5}}$

$\begin{equation*}V=2\pi\int \int \frac{2x-y+4}{\sqrt{5}} dx dy\end{equation}$

Now we just need to work out the bounds.

$0\le y \le 3$ and $0\le x\le \sqrt{12y}$

$\begin{equation*}V=\frac{2\pi}{\sqrt{5}} \int_0^3 \int_0^{\sqrt{12y}} 2x-y+4 dx dy\end{equation}$

$\begin{equation*}V=\frac{2\pi}{\sqrt{5}}\int_0^3 x^2-yx+4x]_0^{\sqrt{12y}} dy \end{equation}$

$\begin{equation*}V=\frac{2\pi}{\sqrt{5}}\int_0^3 12y-\sqrt{12}y^{\frac{3}{2}+4\sqrt{12y} dy\end{equation}$

$\begin{equation*}V=\frac{2\pi}{\sqrt{5}}(6y^2-\frac{2\sqrt{12}}{5}y^{\frac{5}{2}}+\frac{8\sqrt{12}}{3}y^{\frac{3}{2}}]_0^3\end{equation}$

$\begin{equation*}V=\frac{2\pi}{\sqrt{5}}(54-\frac{2\sqrt{12}}{5}(9\sqrt{3})+\frac{8\sqrt{12}}{3}(3\sqrt{3}))\end{equation}$

$\begin{equation*}V=\frac{2\pi}{\sqrt{5}}(54-\frac{108}{5}+48)\end{equation}$

$\begin{equation*}V=\frac{804\pi}{\sqrt{5}}\end{equation}$

If we rationalise the denominator

$\begin{equation*}V=\frac{804\sqrt{5}\pi}{25}\end{equation}$

Leave a Comment

Filed under Integration, Volume of Revolution

Tagged as Volume of revolution, volume of revolution about a slanted line

December 30, 2024 · 8:16 am

Volume of revolution about a line that is not an axis

Find the volume of the solid of revolution obtained by rotating the region bounded by $f(x)=x^3+1, g(x)=x^2, 0\le x\le 1$ about the line $y=3$ .

Rotate the green region about the line $y=3$

Washer Method

$\begin{equation*}V=\pi \int [f(x)]^2 dx \end{equation}$

The volume of the solid is the volume of $y=x^2$ rotated about $y=3$ subtract the volume of $y=x^3+1$ rotated about $y=3$ .

$\begin{equation*}V=\pi \int_0^1((3-x^2)^2-(3-(x^3+1))^2 dx\end{equation}$

$3-x^2$ is the distance (i.e radius) of the curve and the line.

$\begin{equation*}V=\pi \int_0^1(9-6x^2+x^4-(4-4x^3+x^6)) dx\end{equation}$

$\begin{equation*}V=\pi \int_0^1(5-6x^2+4x^3+x^4-x^6) dx\end{equation}$

$\begin{equation*}V=\pi (5x-2x^3+x^4+\frac{x^5}{5}-\frac{x^7}{7}]_0^1\end{equation}$

$\begin{equation*}V=\pi (5-2+1+\frac{1}{5}-\frac{1}{7})\end{equation}$

$\begin{equation*}V=\frac{142 \pi}{35}\end{equation}$

Shell Method

The shell method is much harder because we need to split the integral into two parts.

We need to rotate the green region about $y=3$ and the red region

$\begin{equation*}V=2\pi\int (xf(x))dx\end{equation}$

$\begin{equation*}V=2\pi[\int_0^1(3-y)\sqrt{y} dy+\int_1^2 (3-y)(y-1)^{\frac{1}{3}} dy]\end{equation}$

$3-y$ is the distance between each $y-$ value and the line of rotation. For example, if we were rotating about the $x-$ axis, the distance is $y$ .

$\sqrt{y}$ is the height of the cylinder between $0$ and $1$ . $1-(y-1)^{\frac{1}{3}}$ is the height of the cylinder between $1$ and $2$ . Refer back to Shell method for more information.

I used a calculator to find this integral

Leave a Comment

Filed under Integration, Volume of Revolution, Year 12 Specialist Mathematics

Tagged as Disc method, Volume of revolution, volume of revolution about a line, washer method

December 28, 2024 · 7:12 am

Volume of Revolution Method Two (Shell Method)

I am going to use the same example as I did for Method One (Disc or Washer Method).

If we rotate the shaded region about the $x-$ axis, we get an open hollow cylinder (like a pipe).

The width of the integral is $\delta y$ and the midpoint is $y$ .

The height of the cylinder is $x$ , but we need it in terms of $y$ , hence $x=f(y)$

The volume of the hollow cylinder is the volume of the outer cylinder subtract the volume of the inner cylinder.

$\begin{equation*}V=\pi (y+\frac{\delta y}{2})^2f(y)-\pi (y-\frac{\delta y}{2})^2 f(y)\end{equation}$

$\begin{equation*}V=\pi f(y)((y+\frac{\delta y}{2})^2-(y-\frac{\delta y}{2})^2)\end{equation}$

Which we can expand using a difference of squares.

$\begin{equation*}V=\pi f(y)(y+\frac{\delta y}{2}+y-\frac{\delta y}{2})(y+\frac{\delta y}{2}-y+\frac{\delta y}{2})\end{equation}$

$\begin{equation*}V=\pi f(y)(2y \delta y)\end{equation}$

$\begin{equation*}V=2\pi yf(y)\delta y\end{equation}$

The volume of the entire sold will be

$\begin{equation*}V=\Sigma_{y=a}^b 2 \pi yf(y)\delta y\end{equation}$

As $\delta y \rightarrow 0$

$\begin{equation*}V=\lim\limits_{\delta y \to 0}\Sigma_{y=a}^b 2 \pi yf(y)\delta y=\int_a^b 2\pi yf(y) dy\end{equation}$

Even though we are rotating the line about the $x-$ axis, we are integrating with respect to the $y-$ axis.

Example

Find the volume of the solid generated by revolving the region between $y=x^2$ and $y=2x$ about the $y-$ axis.

If we are rotating about the $y-$ axis, we will integrate with respect to $x$ .

$\begin{equation*}V=2\pi \int x f(x) dx\end{equation}$

The height of our hollow cylinder is $2x-x^2$

Hence

$\begin{equation*}V=2\pi\int_0^2 x(2x-x^2) dx\end{equation}$

$\begin{equation*}V=2\pi \int_0^2 (2x^2-x^3) dx\end{equation}$

$\begin{equation*}V=2\pi (\frac{2x^3}{3}-\frac{x^4}{4}]_0^2\end{equation}$

$\begin{equation*}V=2\pi (\frac{2}{3}\times 8-\frac{1}{4}\times 16 )\end{equation}$

$\begin{equation*}V=32 \pi(\frac{1}{3}-\frac{1}{4})\end{equation}$

$\begin{equation*}V=\frac{8\pi}{3}\end{equation}$

Let’s check with method one.

$x^2=y$ and $x=\frac{y}{2}$

$\begin{equation*}V=\pi \int_0^4 y-\frac{y^2}{4} dy\end{equation}$

$\begin{equation*}V=\pi (\frac{y^2}{2}-\frac{y^3}{12})]_0^4\end{equation}$

$\begin{equation*}V=\pi(8-\frac{16}{3})\end{equation}$

$\begin{equation*}V=\frac{8\pi}{3}\end{equation}$

I try to pick the method that makes the integration easier.

1 Comment

Filed under Integration, Volume of Revolution, Year 12 Specialist Mathematics

Tagged as shell method, Volume of revolution

Differentiating Trigonometric Functions

Trigonometric Limits

HSC Advanced 2024 Question 30

Deriving the Quotient Rule for Differentiation

Example

Deriving the Product Rule for Differentiation

Example

Deriving the Chain Rule for Differentiation

Example

An Imaginary Tale – The story of i

Volume of Rotation About a Slanting Line

Volume of revolution about a line that is not an axis

Volume of Revolution Method Two (Shell Method)

Example

Recent Posts

Categories

Archives