Please Read Me.

DifferentialCalculus

Welcome to Our Site


I greet you this day,
These are the solutions to the SACE past questions on Differential Calculus.
The TI-84 Plus CE shall be used for applicable questions.
The link to the video solutions will be provided for you. Please subscribe to the YouTube channel to be notified of upcoming livestreams. You are welcome to ask questions during the video livestreams.
If you find these resources valuable and if any of these resources were helpful in your passing the Mathematics papers of the SACE, please consider making a donation:

Cash App: $ExamsSuccess or
cash.app/ExamsSuccess

PayPal: @ExamsSuccess or
PayPal.me/ExamsSuccess

Google charges me for the hosting of this website and my other educational websites. It does not host any of the websites for free.
Besides, I spend a lot of time to type the questions and the solutions well. As you probably know, I provide clear explanations on the solutions.
Your donation is appreciated.

Comments, ideas, areas of improvement, questions, and constructive criticisms are welcome.
Feel free to contact me. Please be positive in your message.
I wish you the best.
Thank you.

Difference Quotient and Derivative (Derivatives by Limits)

$ \text{Function} = f(x) \\[3ex] \text{Difference Quotient} = DQ \\[3ex] \text{Derivative} = f'(x) = \dfrac{dy}{dx} = y' \\[5ex] (1.)\;\; DQ = \dfrac{f(x + h) - f(x)}{h} \\[5ex] (2.)\;\; \dfrac{dy}{dx} = \displaystyle{\lim_{h \to 0}} \dfrac{f(x + h) - f(x)}{h} \\[5ex] (3.)\;\; f'(x) = \displaystyle{\lim_{h \to 0}} DQ $

Rules of Derivatives (Derivatives by Rules)

$ a, n \:\:are\:\:constants \\[3ex] (1.)\:\: \underline{Power\;\;Rule} \\[3ex] y = ax^n \\[3ex] \dfrac{dy}{dx} = nax^{n - 1} \\[7ex] (2.)\:\: \underline{Sum/Difference\;\;Rule} \\[3ex] y = u \pm v \pm w \\[3ex] u = f(x);\:\: v = f(x);\;\; w = f(x) \\[3ex] \dfrac{dy}{dx} = \dfrac{du}{dx} \pm \dfrac{dv}{dx} \pm \dfrac{dw}{dx} \\[7ex] (3.)\:\: \underline{Chain\;\;Rule} \\[3ex] y = f(u) \\[3ex] u = f(x) \\[3ex] \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dx} \\[7ex] Also: \\[3ex] y = f(u) \\[3ex] u = f(v) \\[3ex] v = f(x) \\[3ex] \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dv} * \dfrac{dv}{dx} \\[7ex] Also: \\[3ex] y = f(u) \\[3ex] u = f(v) \\[3ex] v = f(w) \\[3ex] w = f(x) \\[3ex] \dfrac{dy}{dx} = \dfrac{dy}{du} * \dfrac{du}{dv} * \dfrac{dv}{dw} * \dfrac{dw}{dx} \\[7ex] ...and\;\;so\;\;on\;\;and\;\;so\;\;forth \\[5ex] (4.)\:\: \underline{Product\;\;Rule} \\[3ex] y = u * v \\[3ex] u = f(x);\:\: v = f(x) \\[3ex] \dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx} \\[7ex] (5.)\:\: \underline{Quotient\;\;Rule} \\[3ex] y = \dfrac{u}{v} \\[5ex] u = f(x);\:\: v = f(x) \\[3ex] \dfrac{dy}{dx} = \dfrac{v\dfrac{du}{dx} - u\dfrac{dv}{dx}}{v^2} $

Standard Derivatives of Exponential Functions

$ a\;\; is\:\:a\;\;positive\:\:constant \\[3ex] k \:\:is\:\:a\;\:constant \\[3ex] (1.)\:\: y = e^x \\[3ex] \dfrac{dy}{dx} = e^x \\[7ex] (2.)\;\; y = e^{kx} \\[3ex] \dfrac{dy}{dx} = ke^{kx} \\[7ex] (3.)\;\; y = e^{-kx} \\[3ex] \dfrac{dy}{dx} = -ke^{kx} \\[7ex] (4.)\:\: y = a^x \\[3ex] \dfrac{dy}{dx} = a^x \ln a \\[7ex] $

Standard Derivatives of Logarithmic Functions

$ a\;\; is\:\:a\;\;positive\:\:constant \\[3ex] k \:\:is\:\:a\;\:constant \\[3ex] (1.)\:\: y = \ln x \\[3ex] \dfrac{dy}{dx} = \dfrac{1}{x} \\[7ex] (2.)\:\: y = \log_a x \\[3ex] \dfrac{dy}{dx} = \dfrac{1}{x \ln a} \\[7ex] (3.)\:\: y = \ln |x| \\[3ex] \dfrac{dy}{dx} = \dfrac{1}{x} \\[7ex] (4.)\:\: y = \log_a |x| \\[3ex] \dfrac{dy}{dx} = \dfrac{1}{x \ln a} \\[7ex] $

Standard Derivatives of Trigonometric Functions

$ (1.)\:\: y = \sin x \\[3ex] \dfrac{dy}{dx} = \cos x \\[7ex] (2.)\:\: y = \cos x \\[3ex] \dfrac{dy}{dx} = -\sin x \\[7ex] (3.)\:\: y = \tan x \\[3ex] \dfrac{dy}{dx} = \sec^2 x \\[7ex] (4.)\:\: y = \csc x \\[3ex] \dfrac{dy}{dx} = -\csc x \cot x \\[7ex] (5.)\:\: y = \sec x \\[3ex] \dfrac{dy}{dx} = \sec x \tan x \\[7ex] (6.)\:\: y = \cot x \\[3ex] \dfrac{dy}{dx} = -\csc^2 x $

Standard Derivatives of Inverse Trigonometric Functions

$ (1.)\:\: y = \sin^{-1} x \\[3ex] \dfrac{dy}{dx} = \dfrac{1}{\sqrt{1 - x^2}} \\[7ex] (2.)\:\: y = \cos^{-1} x \\[3ex] \dfrac{dy}{dx} = \dfrac{-1}{\sqrt{1 - x^2}} \\[7ex] (3.)\;\; y = \tan^{-1} x \\[3ex] \dfrac{dy}{dx} = \dfrac{1}{1 + x^2} \\[7ex] (4.)\:\: y = \csc^{-1} x \\[3ex] \dfrac{dy}{dx} = \dfrac{-1}{|x|\sqrt{x^2 - 1}} \\[7ex] (5.)\:\: y = \sec^{-1} x \\[3ex] \dfrac{dy}{dx} = \dfrac{1}{|x|\sqrt{x^2 - 1}} \\[7ex] (6.)\;\; y = \cot^{-1} x \\[3ex] \dfrac{dy}{dx} = \dfrac{-1}{1 + x^2} \\[7ex] $

Standard Derivatives of Hyperbolic Functions

$ (1.)\:\: y = \sin hx \\[3ex] \dfrac{dy}{dx} = \cos hx \\[7ex] (2.)\:\: y = \cos hx \\[3ex] \dfrac{dy}{dx} = \sin hx \\[7ex] (3.)\;\; y = \tan x \\[3ex] \dfrac{dy}{dx} = \sec^2 hx \\[7ex] (4.)\;\; y = \csc hx \\[3ex] \dfrac{dy}{dx} = -\csc hx \cot hx \\[7ex] (5.)\;\; y = \sec hx \\[3ex] \dfrac{dy}{dx} = -\sec hx \tan hx \\[7ex] (6.)\;\; y = \cot hx \\[3ex] \dfrac{dy}{dx} = -\csc^2 hx \\[7ex] $

Standard Derivatives of Inverse Hyperbolic Functions

$ (1.)\:\: y = \sin h^{-1}x \\[3ex] \dfrac{dy}{dx} = \dfrac{1}{\sqrt{x^2 + 1}} \\[7ex] (2.)\:\: y = \cos h^{-1}x \;\;\;\;\;\;where:\;\; x\gt 1 \\[3ex] \dfrac{dy}{dx} = \dfrac{1}{\sqrt{x^2 - 1}} \\[7ex] (3.)\;\; y = \tan h^{-1}x \;\;\;\;\;\;where:\;\; |x| \lt 1 \\[3ex] \dfrac{dy}{dx} = \dfrac{1}{1 - x^2} \\[7ex] (4.)\;\; y = \csc h^{-1}x \\[3ex] \dfrac{dy}{dx} = \dfrac{-1}{|x|\sqrt{x^2 + 1}} \\[7ex] (5.)\;\; y = \sec h^{-1}x \;\;\;\;\;\;where:\;\; 0\lt x \lt 1 \\[3ex] \dfrac{dy}{dx} = \dfrac{-1}{x\sqrt{1 - x^2}} \\[7ex] (6.)\;\; y = \cot h^{-1}x \;\;\;\;\;\;where:\;\; |x| \gt 1 \\[3ex] \dfrac{dy}{dx} = \dfrac{1}{1 - x^2} \\[7ex] $

Standard Derivatives of Absolute Value Functions

$ (1.)\:\: y = |x| \\[3ex] \dfrac{dy}{dx} = \dfrac{|x|}{x} \\[7ex] $

Newton's Method

$ x_{n + 1} = x_n - \dfrac{f(x)}{f'(x)} $

Mathematical Methods Formula Sheet


Specialist Mathematics Formula Sheet
(1.) Consider the function $f(x) = \dfrac{x}{x - 2}$, where x ≠ 2.

(a.) Use first principles to find $f'(x)$
(b.) Hence, or otherwise, state the slope of the tangent to $f(x)$ when x = 6.


$ (a.) \\[3ex] f(x) = \dfrac{x}{x - 2}\;\;\; x \ne 6 \\[5ex] \text{Difference Quotient, DQ} = \dfrac{f(x + h) - f(x)}{h} \\[5ex] f(x + h) = \dfrac{x + h}{(x + h) - 2} \\[5ex] f(x + h) = \dfrac{x + h}{x + h - 2} \\[5ex] \underline{\text{Numerator of DQ}} \\[3ex] f(x + h) - f(x) \\[3ex] = \dfrac{x + h}{x + h - 2} -\dfrac{x}{x - 2} \\[5ex] = \dfrac{(x + h)(x - 2) - x(x + h - 2)}{(x + h - 2)(x - 2)} \\[5ex] = \dfrac{x^2 - 2x + xh - 2h - x^2 - xh + 2x}{(x + h - 2)(x - 2)} \\[5ex] = \dfrac{-2h}{(x + h - 2)(x - 2)} \\[5ex] \underline{\text{Numerator and Denominator of DQ}} \\[3ex] DQ = [f(x + h) - f(x)] \div h \\[3ex] DQ = \dfrac{-2h}{(x + h - 2)(x - 2)} * \dfrac{1}{h} \\[5ex] DQ = \dfrac{-2}{(x + h - 2)(x - 2)} \\[5ex] \underline{Derivative} \\[3ex] \text{Derivative},\; f'(x) = \displaystyle{\lim_{h \to 0}} DQ \\[3ex] f'(x) = \displaystyle{\lim_{h \to 0}} \dfrac{-2}{(x + h - 2)(x - 2)} \\[5ex] f'(x) = \dfrac{-2}{(x + 0 - 2)(x - 2)} \\[5ex] f'(x) = \dfrac{-2}{(x - 2)(x - 2)} \\[5ex] f'(x) = \dfrac{-2}{(x - 2)^2} \\[5ex] (b.) \\[3ex] \text{slope of the tangent to f(x) when x = 6 is the derivative of f(x) at x = 6} \\[3ex] = \left.\dfrac{dy}{dx}\right|_{x = 6} \\[5ex] = \dfrac{-2}{(6 - 2)^2} \\[5ex] = \dfrac{-2}{4^2} \\[5ex] = \dfrac{-2}{16} \\[5ex] = -\dfrac{1}{8} $
(2.)


(3.)

(4.)

(5.)

(6.)

(7.)


(8.)


(9.)

(10.)


(11.)

(12.)


(13.)

(14.)


(15.)

(16.)


(17.)

(18.)


(19.)

(20.)






Top




(21.)


(22.)


(23.)

(24.)


(25.)

(26.)


(27.)

(28.)


(29.)

(30.)


(31.)

(32.)


(33.)

(34.)


(35.)

(36.)


(37.)

(38.)


(39.)

(40.)


Cash App: Your donation is appreciated. PayPal: Your donation is appreciated. YouTube: Please Subscribe, Share, and Like my Channel
© 2025 Exams Success Group: Your Success in Exams is Our Priority
The Joy of a Teacher is the Success of his Students.