Category Archives: Year 12 Mathematical Methods

May 18, 2026 · 7:24 am

October 28, 2025 · 5:42 am

Integration Question (much easier with Integration by Parts)

The Year 12 Mathematics Methods course doesn’t cover Integration by Parts, so they end up with questions like the following.

Determine the following:
(a) $\frac{d}{dx} \left (e^{2x}sin(3x) \right )$
(b) $\frac{d}{dx} \left (e^{2x}cos(3x) \right )$
Hence, determine the following integral by considering both parts (a) and (b)
$\int_0^{\frac{\pi}{2}}13e^{2x}cos(3x) \enspace dx$

(a) Use the product rule

(1) $\begin{equation*}\frac{d}{dx} \left (e^{2x}sin(3x) \right)= 2e^{2x}sin(3x)+3e^{2x}cos(3x) \end{equation*}$

(b)

(2) $\begin{equation*}\frac{d}{dx} \left (e^{2x}cos(3x) \right )=2e^{2x}cos(3x)-3e^{2x}sin(3x)\end{equation*}$

I need to use equations $1$ and $2$ to find $\int_0^{\frac{\pi}{2}}13e^2xcos(3x) \enspace dx$ .

The $e^{2x}sin(3x)$ terms need to vanish and I need $13$ of the $e^{2x}cos(3x)$ terms.

$3\times \text{equation} (1)+ 2\times \text{equation} (2)$

(3) $\begin{equation*}3\frac{d}{dx} \left (e^{2x}sin(3x) \right)= 6e^{2x}sin(3x)+9e^{2x}cos(3x) \end{equation*}$

(4) $\begin{equation*}2\frac{d}{dx} \left (e^{2x}cos(3x) \right )=4e^{2x}cos(3x)-6e^{2x}sin(3x)\end{equation*}$

Equation $3$ plus equation $4$

(5) $\begin{equation*}3\frac{d}{dx} \left (e^{2x}sin(3x) \right)+2\frac{d}{dx} \left (e^{2x}cos(3x) \right )=13e^{2x}cos(3x)\end{equation*}$

Integrate both sides of the equation

$\int_0^{\frac{\pi}{2}} \left (3\frac{d}{dx} \left (e^{2x}sin(3x) \right)+2\frac{d}{dx} \left (e^{2x}cos(3x) \right ) \right ) dx=\int_0^{\frac{\pi}{2}}13e^{2x}cos(3x) \enspace dx$

By the fundamental theorem of calculus, we know

$(3e^{2x}sin(3x) \right)+2 \left (e^{2x}cos(3x) \right ]_0^{\frac{\pi}{2}}=\int_0^{\frac{\pi}{2}}13e^{2x}cos(3x) \enspace dx$

$3e^{\pi}sin(\frac{3\pi}{2}})+2e^{\pi}cos(\frac{3\pi}{2}})-3e^0sin(3(0))-2e^0cos(3(0))=\int_0^{\frac{\pi}{2}}13e^{2x}cos(3x) \enspace dx$

$\int_0^{\frac{\pi}{2}}13e^{2x}cos(3x) \enspace dx=-3e^{\pi}-2$

Integration by Parts

Remember $\int u\enspace dv=uv-\int v \enspace du$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx$

Let $u=cos(3x)$ , then $du=-3sin(3x)$

and $dv=e^{2x}$ , then $v=\frac{e^{2x}}{2}$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx = \left (cos(3x)\left (\frac{e^{2x}}{2}\right ) \right )_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \frac{e^{2x}}{2}(-3sin(3x)) \enspace dx$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=cos(\frac{3\pi}{2})(\frac{e^\pi}{2})-cos(0)(\frac{e^0}{2})+\frac{3}{2}\int_0^{\frac{\pi}{2}}sin(3x)e^{2x} \enspace dx$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=-\frac{1}{2}+\frac{3}{2}\int_0^{\frac{\pi}{2}}sin(3x)e^{2x} \enspace dx$

Let $u=sin(3x)$ , then $du=3cos(3x)$

and $dv=e^{2x}$ . then $v=\frac{e^{2x}}{2}$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=-\frac{1}{2}+\frac{3}{2}(\frac{e^{2x}}{2}sin(3x)]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}} \frac{e^{2x}}{2}(3cos(3x)) \enspace dx)$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=-\frac{1}{2}+\frac{3}{2}(-\frac{e^\pi}{2}-\frac{3}{2} \int_0^{\frac{\pi}{2}} e^{2x}cos(3x) \enspace dx)$

$\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx=-\frac{1}{2}-\frac{3e^\pi}{4}-\frac{9}{4} \int_0^{\frac{\pi}{2}} e^{2x}cos(3x) \enspace dx$

Collect like terms (the integrals are like)

$\frac{13}{4}(\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx)=-\frac{1}{2}-\frac{3e^\pi}{4}$

$(\int_0^{\frac{\pi}{2}}e^{2x}cos(3x) \enspace dx)=\frac{-2-3e^\pi}{13}$

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Filed under Algebra, Calculus, Integration, Integration by Parts, Year 12 Mathematical Methods

Tagged as integration by parts

October 20, 2025 · 11:11 am

Fundamental Theorem of Calculus

(1) $\begin{equation*}\frac{d}{dx} \left (\int_2^{f(x)} g(t) dt \right )=g(f(x))f'(x)\end{equation*}$

My Year 12 Mathematics Methods students are getting ready for their exam, and questions using the above idea have created a bit of consternation. I am going to work through an example, and show why the ‘formula’ works.

Example

Find $\frac{d}{dx} \left ( \int_2^{x^2} 4t^2+3t \enspace dt \right )$ .

$\begin{equation*}=\frac{d}{dx} \left (\frac{4t^3}{3}+\frac{3t^2}{2} \right ]_2^{x^2}\end{equation}$

$\begin{equation*}=\frac{d}{dx} \left ( \frac{4(x^2)^3}{3}+\frac{3(x^2)^2}{2} -(\frac{4(2^3)}{3}+\frac{3(2^2)}{2} \right )\end{equation}$

$\begin{equation*}=\frac{d}{dx} \left (\frac{4x^6}{3}+\frac{3x^4}{2}-\frac{50}{3} \right) \end{equation}$

$\begin{equation*}=\frac{24x^5}{3}+\frac{12x^3}{2}\end{equation}$

(2) $\begin{equation*}=8x^5+6x^3\end{equation*}$

If we used ‘formula’ $1$

$\begin{equation*}\frac{d}{dx} \left ( \int_2^{x^2} 4t^2+3t \enspace dt \right )=(4(x^2)^2+3(x^2))(2x)\end{equation}$

(3) $\begin{equation*}\frac{d}{dx} \left ( \int_2^{x^2} 4t^2+3t \enspace dt \right )=8x^5+6x^3 \end{equation*}$

We can see equation $2$ and $3$ are the same.

More formally

$\begin{equation*}\frac{d}{dx} \left ( \int_c^{f(x)} g(t) dt \right )=\frac{d}{dx} \left (G(f(x))-G(c) \right ) \end{equation}$

Remember $\frac{d}{dx} \left (f(g(x)) \right )=f'(g(x))\times g'(x)$

$\begin{equation*}\frac{d}{dx} \left (G(f(x))-G(c) \right ) =G'(f(x))f'(x)=g(f(x))\times f'(x)\end{equation}$

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

Tagged as differentiation, fundamental theorem of calculus

July 22, 2025 · 6:43 am

Continuous Uniform Random Variable

My Year 12 Mathematics Methods students are doing continuous random variables at the moment and I thought it would be worthwhile deriving the mean and variance formulas for a uniform continuous random variable.

The probability density function for a uniform random variable is

$\begin{equation*}f(x)= \left \{ {\begin{matrix}\frac{1}{b-a} & a\le x\le b \\0 & \text {elsewhere}\end{matrix}}\end{equation}$

and it looks like

Remember, the mean $\mu$ or expected value $E(X)$ of a continuous random variable is

(1) $\begin{equation*}E(X)=\int xp(x) dx\end{equation*}$

and the variance $\sigma^2$ is

(2) $\begin{equation*}\sigma^2=\int (x-\mu)^2p(x) dx\end{equation*}$

We are going to use equations $1$ and $2$ to find formulae for a uniform continuous random variable.

$\begin{equation*}\mu=\int_a^b x (\frac{1}{b-a}) dx\end{equation}$

$\begin{equation*}\mu=\frac{x^2}{2(b-a)}|\begin{matrix}b\\a\end{matrix}\end{equation}$

$\begin{equation*}\mu=\frac{b^2}{2(b-a)}-\frac{a^2}{2(b-a)}=\frac{b^2-a^2}{2(b-a)}\end{equation}$

Factorise the numerator (using difference of squares)

$\begin{equation*}\mu=\frac{(b-a)(b+a)}{2(b-a)}\end{equation}$

Hence,

$\begin{equation*}\mu=\frac{b+a}{2}\end{equation}$

Now for the variance

$\begin{equation*}\sigma^2=\int_a^b (x-(\frac{a+b}{2}))^2(\frac{1}{b-a}) dx\end{equation}$

$\begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{(x-\frac{a+b}{2})^3}{3})|\begin{matrix}b\\a\end{matrix}\end{equation}$

$\begin{equation*}\sigma^2=\frac{1}{b-a}((\frac{(b-\frac{a+b}{2})^3}{3})-(\frac{(a-\frac{a+b}{2})^3}{3}))\end{equation}$

$\begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{-a^3}{12}+\frac{b^3}{12}+\frac{a^2b}{4}-\frac{ab^2}{4})\end{equation}$

$\begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{-a^3+b^3+3a^2b-3ab^2}{12})\end{equation}$

$\begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{b^3-3b^2a+3ba^2-a^3}{12})\end{equation}$

From the binomial expansion theorem, we know

$\begin{equation*}b^3-3b^2a+3ba^2-a^3=(b-a)^3\end{equation}$

Hence

$\begin{equation*}\sigma^2=\frac{1}{b-a}(\frac{(b-a)^3}{12}\end{equation}$

and

$\begin{equation*}\sigma^2=\frac{(b-a)^2}{12}\end{equation}$

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Filed under Binomial Expansion Theorem, Continuous Random Variables, Probability Distributions, Uniform, Year 12 Mathematical Methods

Tagged as mean, uniform continuous random variable, variance

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

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

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Filed under Calculus, Differentiation, Year 12 Mathematical Methods

Tagged as differentiating e

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

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

Tagged as differentiating tan, differentiating trig

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

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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.

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

Tagged as trig limits, trigonometric limits

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.

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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.

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

Category Archives: Year 12 Mathematical Methods

Using Integration to find the Centroid of an Area

Integration Question (much easier with Integration by Parts)

Integration by Parts

Fundamental Theorem of Calculus

Example

More formally

Continuous Uniform Random Variable

Differentiating f(x)=e^x

Differentiating the Tangent Function

Differentiating Trigonometric Functions

Trigonometric Limits

Deriving the Quotient Rule for Differentiation

Example

Deriving the Product Rule for Differentiation

Example

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