Category Archives: Year 12 Mathematical Methods

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

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.

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

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

November 2, 2024 · 11:45 am

Effect of Function Transformations on Integration

My year 12 Mathematical Methods students have questions like this

Given that $f(x)$ is continuous everywhere and that $\int_{4}^{10} f(x) dx=-10$ , find:

(a) $\int_{4}^{10}2x +f(x) dx$

(b) $\int_{5}^{11} f(x-1) dx$

(c) $\int_{1}^{3} f(3x+1) dx$

(d) $\int_{-10}^{-4} -f(-x) dx$

(e) $\int{10}^{22} f(\frac{x-2}{2}) dx$

(f) $\int_{-3}^{-9} f(1-x) dx$

OT Lee Mathematics Methods Textbook Ex 8.3 question 6

For the most part these questions aren’t too difficult, but the horizontal dilations cause issues.

\int_{4}^{10} 2x dx +\int_{2}^{10} f(x) dx

Once you get the hang of it, you can skip the change of variable and multiply the value of the definite integral by the scale factor of the horizontal dilation (only if the integration bounds are also changed).