Category Archives: 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).