Differential Calculus

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These are the solutions to the CSEC Additional Mathematics past questions on Differential Calculus.
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Rules of Derivatives

$ 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 \\[3ex] (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)} $

Formula Sheet: List of Formulae

(1.) (a.) (i.) Use the definition of the derivative as a limit to find $f'(x)$ for the function $f(x) = x^2 - 3$

(ii.) Hence, determine the value of $f''(5)$

(b.) Given that $y = \dfrac{\sin x}{\cos x}$, show that $\dfrac{dy}{dx} = \dfrac{1}{\cos^2 x}$

(c.) The equation of a curve is given by $y = (5x^2 - 7)^4$
Determine the equation of the gradient of the curve.

$ (a.) (i.) \\[3ex] \underline{\text{Derivative by Limits}} \\[3ex] f(x) = x^2 - 3 \\[4ex] f(x + h) = (x + h)^2 - 3 \\[4ex] = (x + h)(x + h) - 3 \\[3ex] = x^2 + xh + xh + h^2 - 3 \\[4ex] = x^2 + 2xh + h^2 - 3 \\[5ex] f(x + h) - f(x) = x^2 + 2xh + h^2 - 3 - (x^2 - 3) \\[4ex] = x^2 + 2xh + h^2 - 3 - x^2 + 3 \\[4ex] = 2xh + h^2 \\[4ex] = h(2x + h) \\[5ex] \dfrac{f(x + h) - f(x)}{h} = \dfrac{h(2x + h)}{h} \\[5ex] = 2x + h \\[5ex] f'(x) = \displaystyle{\lim_{h \to 0}}\:\:2x + h \\[4ex] f'(x) = 2x + 0 \\[3ex] f'(x) = 2x \\[5ex] (ii.) \\[3ex] \underline{\text{Derivative by Rules}} \\[3ex] f'(x) = 2x \\[3ex] f''(x) = 2 \\[3ex] f''(5) = 2 \\[5ex] (b.) \\[3ex] y = \dfrac{\sin x}{\cos x} \\[5ex] \underline{\text{Quotient Rule}} \\[3ex] u = \sin x \hspace{4em} v = \cos x \\[3ex] u' = \cos x \hspace{4em} v' = -\sin x \\[5ex] \dfrac{dy}{dx} = \dfrac{vu' - uv'}{v^2} \\[6ex] = \dfrac{\cos x(\cos x) - \sin x(-\sin x)}{\cos^2 x} \\[6ex] = \dfrac{\cos^2 x + \sin^2 x}{\cos^2 x} \\[5ex] \cos^2 x + \sin^2 x = 1 ...\text{Pythagorean Identity} \\[4ex] = \dfrac{1}{\cos^2 x} \\[6ex] $ The equation of the gradient of the curve is the derivative of the curve

$ (c.) \\[3ex] y = (5x^2 - 7)^4 \\[4ex] \underline{\text{Chain Rule}} \\[3ex] Let\;\;p = 5x^2 - 7 \hspace{4em} y = p^4 \\[4ex] \dfrac{dp}{dx} = 10x \hspace{5em} \dfrac{dy}{dp} = 4p^3 \\[6ex] \dfrac{dy}{dx} = \dfrac{dy}{dp} * \dfrac{dp}{dx} \\[5ex] = 4p^3 * 10x \\[4ex] = 40xp^3 \\[4ex] = 40x(5x^2 - 7)^3 $

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