Monthly Archives: February 2025

February 25, 2025 · 9:28 am

Differentiating f(x)=e^x

We are going to differentiate $f(x)=e^x$ from first principals.

Remember the definition of a derivative is

(1) $\begin{equation*}f'(x)=\lim_{\limits{h\to 0}}\frac{f(x+h)-f(x)}{h}\end{equation*}$

If $f(x)=e^x$ , then

$\begin{equation*}f'(x)=\lim_{\limits{h \to 0}}\frac{e^{x+h}-e^x}{h}\end{equation}$

$\begin{equation*}f'(x)=\lim_{\limits{h \to 0}}\frac{e^x\times e^h-e^x}{h}\end{equation}$

$\begin{equation*}f'(x)=\lim_{\limits{h \to 0}}\frac{e^x(e^h-1)}{h}\end{equation}$

$\begin{equation*}f'(x)=e^x\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}\end{equation}$

Let’s think about $\lim_{\limits{h \to 0}}\frac{e^h-1}{h}$

Remember $e$ is defined as $\lim_{\limits{n \to \infty}}(1+\frac{1}{n})^n$

(2) $\begin{equation*}\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}\end{equation*}$

Let $y=e^h-1$ , as $h \to 0, e^h-1 \to 1-1=0$ hence $y \to 0$

If $y=e^h-1$ , then $h=ln(y+1)$

(3) $\begin{equation*}\lim_{\limits{y \to 0}}\frac{y}{ln(y+1)}\end{equation*}$

We are going to rewrite the equation $3$ as

(4) $\begin{equation*}\lim_{\limits{y \to 0}}\frac{\frac{1}{ln(y+1)}}{y}\end{equation*}$

And then we can write equation $4$ as

(5) $\begin{equation*}{\lim_{\limits{y \to 0}}{\frac{1}{\frac{1}{y}ln(y+1)}\end{equation*}$

Using log laws we can write equation $5$ as

(6) $\begin{equation*}\lim_{\limits{y \to 0}}\frac{1}{ln(y+1)^{\frac{1}{y}}}\end{equation*}$

Let $y=\frac{1}{n}$

As $y \to 0, n \to \infty$

equation $6$ becomes

(7) $\begin{equation*}\lim_{\limits{n \to \infty}}\frac{1}{ln(\frac{1}{n}+1)^n}\end{equation*}$

We can move the limit to inside the natural log

(8) $\begin{equation*}\frac{1}{ln(\lim_{\limits{n \to \infty}}(\frac{1}{n}+1)^n)}\end{equation*}$

And we know from the definition of $e$ that $e=\lim_{\limits{n \to \infty}}(1+\frac{1}{n})^n$

Hence, equation $8$ is

(9) $\begin{equation*}\frac{1}{ln(e)}=1\end{equation*}$

Back to our derivative

$\begin{equation*}f'(x)=e^x\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}\end{equation}$

We know that $\lim_{\limits{h \to 0}}{\frac{e^h-1}{h}=1$ hence

$\begin{equation*}f'(x)=e^x\end{equation}$

Finding a Recursive Rule for a Sequence (Year 12 Mathematics Applications)

How do we go about finding the rule for a first order linear recurrence relation?

Something like

$\begin{equation*}5, 7, 11, 19, ...\end{equation}$

There isn’t a common difference (arithmetic sequence) or a common ratio (geometric sequence). Sometimes you can just see the rule, but an algorithm will be handy.

Let’s say our relationship is

(1) $\begin{equation*}T_{n+1}=bT_n+c, T_1=a\end{equation*}$

Referring back to our sequence $5, 7, 11, 19, ...$ , we know

(2) $\begin{equation*}7=b\times5+c\end{equation*}$

and

(3) $\begin{equation*}11=b\times7+c\end{equation*}$

We can solve equation $2$ and $3$ simultaneously

equation $2-$ equation $3$

$\begin{equation*}-4=-2b\end{equation}$

Hence $b=2$

Substitute $b=2$ into equation $2$

$\begin{equation*}7=2\times 5+c\end{equation}$

$\begin{equation*}7=10+c\end{equation}$

Hence $c=-3$

$\begin{equation*}T_{n+1}=2T_n-3, T_1=5\end{equation}$

Let’s try to generalise

If $T_{n+1}=bT_n+c, T_1=a$ , then

(4) $\begin{equation*}T_2=bT_1+c\end{equation*}$

and

(5) $\begin{equation*}T_3=bT_2+c\end{equation*}$

Equation $5 -$ equation $4$

$\begin{equation*}T_3-T_2=(T_2-T_1)b\end{equation}$

$\begin{equation*} b=\frac{T_3-T_2}{T_2-T_1}\end{equation}$

Hence, $b=\frac{T_{n+2}-T_{n+1}}{T_{n+1}-T_n}$

Once you know $b$ , substitute into either equation to find $C$ .

Example

Find the recursive rule for the following

$-8, -12, -20, -36, ...$

$\begin{equation*}b=\frac{T_{n+2}-T_{n+1}}{T_{n+1}-T_n}=\frac{-20-(-12)}{-12-(-8)}=\frac{-8}{-4}=2\end{equation}$

$\begin{equation*}-12=2\times-8+c\end{equation}$

$\begin{equation*}-12=-16+c\end{equation}$

Hence $c=4$ and $T_{n+1}=2T_n+4, T_1=-8$

It is also possible to find the rule using a Classpad (if it’s in the calculator section} by using an e-activity.

Counting Techniques

Four teachers decide to swap desks at work. How many ways can this be done if no teacher sits at their previous desk?
Mathematics Specialist Units 1&2 Cambridge

I like this question as it seems easy until you start thinking about it. I think the best approach is a tree diagram.

If we think of the four teachers as A, B, C and D. Then A can no longer sit in A, so the options are B, C and D for the first desk.

For the second desk, If B is in the first desk, then A, C or D could be in the second. If C is in the first desk, then A or D could be in the second (B can’t be in the same desk). If D is in the first desk, then A or C can be in the second desk.

And so on, leaving 9 possibilities

BADC
BCDA
BDAC
CADB
CDAB
CDBA
DABC
DCAB
DCBA

1 Comment

Filed under Counting Techniques, Tree Diagram, Year 11 Specialist Mathematics

Tagged as counting techniques, Year 11 Specialist Mathematics

February 4, 2025 · 9:17 am

Differentiating the Tangent Function

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

I use the quotient rule to differentiate $f(x)=tan(x)$ .

(1) $\begin{equation*}\frac{d}{dx}(\frac{f(x)}{g(x)})=\frac{f'(x)g(x)-g'(x)f(x)}{[g(x)]^2}\end{equation*}$

If $h(x)=tan(x)=\frac{sin(x)}{cos(x)}$ then from equation $1$

(2) $\begin{equation*}h'(x)=\frac{cos(x)cos(x)-(-sin(x)sin(x))}{[cos(x)]^2}\end{equation*}$

(3) $\begin{equation*}h'(x)=\frac{cos^2(x)+sin^2(x)}{cos^2(x)}\end{equation*}$

Remember the Pythagorean identity

(4) $\begin{equation*}sin^2(x)+cos^2(x)=1\end{equation*}$

Hence

$\begin{equation*}h'(x)=\frac{1}{cos^2(x)}=sec^2(x)\end{equation}$

(5) $\begin{equation*}\frac{d}{dx}tan(x)=sec^2(x)\end{equation*}$

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

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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

Monthly Archives: February 2025

Differentiating f(x)=e^x

Finding a Recursive Rule for a Sequence (Year 12 Mathematics Applications)

Example

Counting Techniques

Differentiating the Tangent Function

Differentiating Trigonometric Functions

Trigonometric Limits

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