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Differential Calculus

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These are the solutions to Mathematics questions on Differential Calculus.
The TI-84 Plus CE shall be used for applicable questions.
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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)} $

(1.) Find the derivative

$ (i.)\;\; y = x^5\sin^2 x\cos 4x \\[3ex] (ii.)\;\; y = xy - 2x^2y^2 - 5x \\[3ex] (iii.)\;\; y = (x^2 - 1)(3 - 2x)^2 \\[3ex] $

$ (i.) \\[3ex] y = x^5\sin^2 x\cos 4x \\[3ex] y = x^5 * \sin^2 x * \cos 4x \\[3ex] y = u * v * w \\[3ex] Let\;\;u = x^5 \hspace{3em} v = \sin^2 x \hspace{3em} w = \cos 4x \\[3ex] .................................................................... \\[3ex] u = x^5 \\[3ex] u' = 5x^4...\text{Power Rule} \\[5ex] v = \sin^2 x \\[3ex] Let\;\; m = \sin x \hspace{2em}\implies\hspace{2em} v = m^2 \\[3ex] \dfrac{dm}{dx} = \cos x \hspace{5em} \dfrac{dv}{dm} = 2m \\[5ex] \dfrac{dv}{dx} = \dfrac{dv}{dm} * \dfrac{dm}{dx} ...\text{Function of a Function Rule} \\[5ex] v' = 2m * \cos x \\[3ex] v' = 2\sin x \cos x \\[5ex] w = \cos 4x \\[3ex] Let\;\; n = 4x \hspace{2em}\implies\hspace{2em} w = \cos n \\[3ex] \dfrac{dn}{dx} = 4 \hspace{5em} \dfrac{dw}{dn} = -\sin n \\[5ex] \dfrac{dw}{dx} = \dfrac{dw}{dn} * \dfrac{dn}{dx} ...\text{Function of a Function Rule} \\[5ex] w' = -\sin n * 4 \\[3ex] w' = -4\sin 4x \\[3ex] .................................................................... \\[3ex] y' = u'vw + uv'w + uvw' ...\text{Product Rule} \\[3ex] = 5x^4(\sin^2 x)(\cos 4x) + x^5(2\sin x \cos x)(\cos 4x) + x^5(\sin^2 x)(-4\sin 4x) \\[3ex] = 5x^4\sin^2 x\cos 4x + 2x^5\sin x\cos x\cos 4x - 4x^5\sin^2x\sin 4x \\[3ex] = x^4\sin x[5\sin x\cos 4x + 2x\cos x\cos 4x - 4x\sin x\sin 4x] \\[3ex] = x^4\sin x[\cos 4x(5\sin x + 2x\cos x) - 4x\sin x\sin 4x] $

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