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] $

(2.) (a.) Evaluate the $\displaystyle{\lim_{x \to \infty}} \dfrac{3x}{x + 2}$

(b.) Find the derivative of $y = 5 + 7x + 3x^2$

$ (a.) \\[3ex] \displaystyle{\lim_{x \to \infty}} \dfrac{3x}{x + 2} \\[5ex] = \displaystyle{\lim_{x \to \infty}} \dfrac{\dfrac{3x}{x}}{\dfrac{x}{x} + \dfrac{2}{x}} \\[10ex] = \displaystyle{\lim_{x \to \infty}} \dfrac{3}{1 + \dfrac{2}{x}} \\[8ex] = \dfrac{3}{1 + 0} \\[5ex] = \dfrac{3}{1} \\[5ex] = 3 \\[5ex] (b.) \\[3ex] y = 5 + 7x + 3x^2 \\[3ex] \dfrac{dy}{dx} = 0 + 7 + 6x \\[5ex] = 6x + 7 $

(3.) Differentiate these functions with respect to (wrt) x

$ f(x) = \left(\dfrac{x + 2}{x + 4}\right)^9 \\[5ex] $

$ f(x) = \left(\dfrac{x + 2}{x + 4}\right)^9 \\[5ex] y = \left(\dfrac{x + 2}{x + 4}\right)^9 \\[5ex] \text{Let }p = \dfrac{x + 2}{x - 4} \hspace{3em} \implies y = p^9 \\[5ex] ........................................................... \\[3ex] p = \dfrac{x + 2}{x - 4} = \dfrac{u}{v} \\[5ex] \text{Let }u = x + 2 \hspace{5em} v = x - 4 \\[3ex] u' = \dfrac{du}{dx} = 1 \hspace{6em} v' = \dfrac{dv}{dx} = 1 ...\text{Power Rule} \\[5ex] \dfrac{dp}{dx} = \dfrac{vu' - uv'}{v^2}...\text{Quotient Rule} \\[5ex] = \dfrac{(x - 4)(1) - (x + 2)(1)}{(x - 4)^2} \\[5ex] = \dfrac{x - 4 - x - 2}{(x - 4)^2} \\[5ex] = -\dfrac{6}{(x - 4)^2} \\[5ex] ........................................................... \\[3ex] y = p^9 \\[3ex] \dfrac{dy}{dp} = 9p^8 ...\text{Power Rule} \\[3ex] ........................................................... \\[3ex] \dfrac{dy}{dx} = \dfrac{dy}{dp} * \dfrac{dp}{dx} ...\text{Function of a Function Rule} \\[5ex] = 9p^8 * -\dfrac{6}{(x - 4)^2} \\[5ex] = -54p^8 \div (x - 4)^2 \\[3ex] = -54\left(\dfrac{x + 2}{x - 4}\right)^8 * \dfrac{1}{(x - 4)^2} \\[5ex] = -\dfrac{54(x + 2)^8}{(x - 4)^8 * (x - 4)^2} \\[5ex] = -\dfrac{54(x + 2)^8}{(x - 4)^{10}} $

(4.) Consider a function $y = 3x^2 + 2x - 7$
(a.) Derive the gradient function $\dfrac{dy}{dx}$
(b.) Use it to find the local minimum or maximum point.
Assess whether it is a minimum or a maximum.

$ y = 3x^2 + 2x - 7 \\[3ex] f(x) = 3x^2 + 2x - 7 \\[3ex] (a.) \\[3ex] \dfrac{dy}{dx} = 6x + 2 \\[5ex] (b.) \\[3ex] \underline{\text{Second Derivative Test}} \\[3ex] \text{Critical Points: Set the derivative to zero and solve for }x \\[3ex] 6x + 2 = 0 \\[3ex] 6x = -2 \\[3ex] x = -\dfrac{2}{6} = -\dfrac{1}{3} \\[5ex] \text{Find the function value at the critical point} \\[3ex] f\left(-\dfrac{1}{3}\right) \\[5ex] = 3\left(-\dfrac{1}{3}\right)^2 + 2\left(-\dfrac{1}{3}\right) - 7 \\[5ex] = 3\left(-\dfrac{1}{9}\right) - \dfrac{2}{3} - \dfrac{21}{3} \\[5ex] = \dfrac{1}{3} - \dfrac{2}{3} - \dfrac{21}{3} \\[5ex] = \dfrac{1 - 2 - 21}{3} \\[5ex] = -\dfrac{22}{3} \\[5ex] \text{Point } \left(-\dfrac{1}{3}, -\dfrac{22}{3}\right) \text{ is a local maximum or a local minimum} \\[5ex] \text{To determine if the point is a local maximum or a local minimum:} \\[3ex] \text{Second Derivative: Find the derivative of the derivative} \\[3ex] \dfrac{d^2y}{dx^2} = 6 \\[5ex] 6 \gt 0 \\[3ex] \text{Hence, Point } \left(-\dfrac{1}{3}, -\dfrac{22}{3}\right) \text{ is a local minimum} \\[5ex] \text{To confirm if the point is a local maximum or a local minimum:} \\[3ex] \text{Critical Value, } x = -\dfrac{1}{3} \\[5ex] \text{Test Intervals are: } x \lt -\dfrac{1}{3} \hspace{1em}\text{and}\hspace{1em} x \gt -\dfrac{1}{3} \\[5ex] \text{First Derivative Test: Sign Test} \\[3ex] $

$\dfrac{dy}{dx} = 6x + 2$
$x \lt -\dfrac{1}{3}$ Let x = −1	$x \gt -\dfrac{1}{3}$ Let x = 0
$ 6(-1) + 2 \\[3ex] -4 \\[3ex] $	$ 6(0) + 2 \\[3ex] 2 \\[3ex] $
negative	positive
Because the slope changes from negative to positive, the function changes from decreasing to increasing. This implies that $x = -\dfrac{1}{3}$ is a local minimum.

(5.) You have a pond with fish.
You feed the fish with f kg fish food monthly.
You have noticed that the number N of fish living in your pond depends on the amount of feed.
You have even worked out a formula: $N = -f^2 + 8f + 5$

(a.) According to this function, if you stop feeding your fish, will they all die?
(b.) What amount of feed will give you the greatest amount of fish?
(c.) What is the greatest amount of fish at any given time?
(d.) Calculate the gradient function $\dfrac{dN}{df}$ for $f = 2$.
Explain what this value tells you about the development of the number of your fish.

$ N(f) = -f^2 + 8f + 5 \\[3ex] (a.) \\[3ex] f = 0 ...\text{no more feed for the fish} \\[3ex] N(0) = -0^2 + 8(0) + 5 \\[3ex] N(0) = 5 \\[3ex] $ According to this function, if you stop feeding the fish, they will not all die.
5 fish will remain.

The function is a quadratic function, so let us find the vertex.
The amount of feed will give you the greatest amount of fish is the x-coordinate of the vertex.
The greatest amount of fish is the y-coordinate of the vertex.

$ N(f) = -f^2 + 8f + 5 \\[3ex] \text{Compare to the standard form: } N(f) = af^2 + bf + c \\[3ex] a = -1 \\[3ex] b = 8 \\[3ex] c = 5 \\[3ex] x-coordinate = -\dfrac{b}{2a} \\[5ex] = -\dfrac{8}{2(-1)} \\[5ex] = 4\;kg \\[5ex] y-coordinate = N(4) \\[3ex] = -4^2 + 8(4) + 5 \\[3ex] = 21\;fish \\[3ex] $ (b.) The amount of feed that gives the greatest amount of fish is 4kg per month.
(c.) The greatest amount of fish at any given time is 21 fish.

$ (d.) \\[3ex] N(f) = -f^2 + 8f + 5 \\[3ex] \dfrac{dN}{df} = -2f + 8 \\[5ex] \left.\dfrac{dN}{df}\right|_{f = 2} \\[5ex] = -2(2) + 8 \\[3ex] = 4 \\[3ex] $ The instantaneous rate of change in the number of fish when the amount of feed is 2 kg, is 4 fish per kilogram of feed.
This implies that at the feeding level of 2 kg per feed monthly, each additional kilogram of feed would increase the fish population at a rate of about 4 fish.

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