Differential Calculus

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These are the solutions to the SACE past questions on Differential Calculus.
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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} $

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