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.
$ \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 $
$ 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} $
$ 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] $
$ 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] $
$ (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 $
$ (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] $
$ (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] $
$ (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] $
$ (1.)\:\: y = |x| \\[3ex] \dfrac{dy}{dx} = \dfrac{|x|}{x} \\[7ex] $
$ x_{n + 1} = x_n - \dfrac{f(x)}{f'(x)} $
© 2025 Exams Success Group:
Your
Success in Exams is Our Priority
The Joy of a Teacher is the Success of his
Students.