Solved Examples on Derivatives and the Applications of Derivatives

We can solve this question in two ways.
Use any method you prefer.

$ \underline{First\:\:Method:\:\:By\:\:Definition} \\[3ex] f(x) = x^2 - 5 \\[3ex] f(x + \Delta x) = (x + \Delta x)^2 - 5 \\[3ex] = (x + \Delta x)(x + \Delta x) - 5 \\[3ex] = x^2 + x\Delta x + x\Delta x + (\Delta x)^2 - 5 \\[3ex] = x^2 + 2x\Delta x + (\Delta x)^2 - 5 \\[5ex] f(x + \Delta x) - f(x) \\[3ex] = x^2 + 2x\Delta x + (\Delta x)^2 - 5 - (x^2 - 5) \\[3ex] = x^2 + 2x\Delta x + (\Delta x)^2 - 5 - x^2 + 5 \\[3ex] = 2x\Delta x + (\Delta x)^2 \\[3ex] = \Delta x(2x + \Delta x) \\[5ex] \dfrac{f(x + \Delta x) - f(x)}{\Delta x} \\[5ex] = \dfrac{\Delta x(2x + \Delta x)}{\Delta x} \\[5ex] = 2x + \Delta x \\[3ex] f'(x) = \displaystyle{\lim_{\Delta x \to 0}}\:\:2x + \Delta x \\[3ex] f'(x) = 2x + 0 \\[3ex] f'(x) = 2x \\[5ex] \underline{Second\:\:Method:\:\:By\:\:Formula} \\[3ex] f'(x) = \displaystyle{\lim_{h \to 0}}\: \dfrac{f(x + h) - f(x)}{h} \\[5ex] f(x) = x^2 - 5 \\[3ex] f(x + h) = (x + h)^2 - 5 \\[3ex] = (x + h)(x + h) - 5 \\[3ex] = x^2 + hx + hx + h^2 - 5 \\[3ex] = x^2 + 2xh + h^2 - 5 \\[3ex] \underline{Numerator} \\[3ex] f(x + h) - f(x) \\[3ex] = x^2 + 2xh + h^2 - 5 - (x^2 - 5) \\[3ex] = x^2 + 2xh + h^2 - 5 - x^2 + 5 \\[3ex] = 2xh + h^2 \\[3ex] = h(2x + h) \\[3ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{h(2x + h)}{h} \\[5ex] = 2x + h \\[3ex] f'(x) = \displaystyle{\lim_{h \to 0}}\: 2x + h \\[3ex] f'(x) = 2x + 0 \\[3ex] f'(x) = 2x $

Power Rule and Difference Rule

$ f(x) = x^2 - 5 \\[3ex] f'(x) = 2 * x^{2 - 1} - 5x^0 \\[3ex] = 2 * x^1 - 0 * 5 * x^{0 - 1} \\[3ex] = 2 * x - 0 \\[3ex] f'(x) = 2x $

The maximum value is the $y-value$
The derivative is used to find the maximum value.
We shall use it as our first method.
First, find the derivative first. Second, set it to zero and solve for $x$. Third, substitute the value of $x$ in the main equation in order to find the $y$
However, the graph is a parabola. So, we can also find the vertex of the parabola.
That shall be our second method.
Use whichever method you prefer - that you feel is fast and accurate.

$ \underline{First\:\:Method - Derivatives} \\[3ex] y = 1 - 2x - 3x^2 \\[3ex] \dfrac{dy}{dx} = -2 - 6x \\[5ex] Set\:\:\dfrac{dy}{dx} = 0 \\[5ex] -2 - 6x = 0 \\[3ex] -2 = 6x \\[3ex] 6x = -2 \\[3ex] x = -\dfrac{2}{6} \\[5ex] x = -\dfrac{1}{3} \\[5ex] Substitute\:\:for\:\:x\:\:in\:\:the\:\:main\:\:equation \\[3ex] y = 1 - 2\left(-\dfrac{1}{3}\right) - 3\left(-\dfrac{1}{3}\right)^2 \\[5ex] y = 1 + \dfrac{2}{3} - 3\left(-\dfrac{1}{3}\right)\left(-\dfrac{1}{3}\right) \\[5ex] y = \dfrac{3}{3} + \dfrac{2}{3} - \dfrac{1}{3} \\[5ex] y = \dfrac{3 + 2 - 1}{3} \\[5ex] y = \dfrac{5}{3} \\[5ex] \underline{Second\:\:Method - Vertex\:\:Formula} \\[3ex] y = 1 - 2x - 3x^2 \\[3ex] y = -3x^2 - 2x + 1 \\[3ex] y = ax^2 + bx + c \\[3ex] Compare \\[3ex] a = -3 \\[3ex] b = -2 \\[3ex] Vertex = \left(-\dfrac{b}{2a}, f\left(-\dfrac{b}{2a}\right)\right) \\[5ex] x = -\dfrac{b}{2a} = \dfrac{-(-2)}{2(-3)} = \dfrac{2}{-6} = -\dfrac{2}{6} = -\dfrac{1}{3} \\[5ex] Substitute\:\:for\:\:x\:\:in\:\:the\:\:equation \\[3ex] y = 1 - 2\left(-\dfrac{1}{3}\right) - 3\left(-\dfrac{1}{3}\right)^2 \\[5ex] y = 1 + \dfrac{2}{3} - 3\left(-\dfrac{1}{3}\right)\left(-\dfrac{1}{3}\right) \\[5ex] y = \dfrac{3}{3} + \dfrac{2}{3} - \dfrac{1}{3} \\[5ex] y = \dfrac{3 + 2 - 1}{3} \\[5ex] y = \dfrac{5}{3} $

The gradient(slope) of the tangent line is the derivative

$ y = 2kx^2 + x + 1 \:\:at\:\:x = 1 \\[3ex] \dfrac{dy}{dx} = 4kx + 1 \\[5ex] At\:\: x = 1, \:\:\dfrac{dy}{dx} = 9 \\[5ex] \dfrac{dy}{dx}\Big|_{x = 1} = 4 * k * 1 + 1 = 9 \\[5ex] 4k + 1 = 9 \\[3ex] 4k = 9 - 1 \\[3ex] 4k = 8 \\[3ex] k = \dfrac{8}{4} \\[5ex] k = 2 $

We can solve this question in two ways.
Use any method you prefer.

$ \underline{First\:\:Method:\:\:By\:\:Definition} \\[3ex] y = \dfrac{1}{x}...eqn.(1) \\[5ex] y + \Delta y = \dfrac{1}{x + \Delta x} \\[5ex] Subtract\:\:eqn.(1) \\[3ex] y + \Delta y - y = \dfrac{1}{x + \Delta x} - \dfrac{1}{x} \\[5ex] \Delta y = \dfrac{1(x) - 1(x + \Delta x)}{x(x + \Delta x)} \\[5ex] \Delta y = \dfrac{x - x - \Delta x}{x(x + \Delta x)} \\[5ex] \Delta y = -\dfrac{\Delta x}{x(x + \Delta x)} \\[5ex] Divide\:\:both\:\:sides\:\:by\:\:\Delta x \\[3ex] \dfrac{\Delta y}{\Delta x} = \dfrac{1}{\Delta x} * -\dfrac{\Delta x}{x(x + \Delta x)} \\[5ex] \dfrac{\Delta y}{\Delta x} = -\dfrac{1}{x(x + \Delta x)} \\[5ex] Limit\:\:as\:\:\Delta x\rightarrow 0 \\[3ex] \displaystyle{\lim_{\Delta x \to 0}} \dfrac{\Delta y}{\Delta x} = \displaystyle{\lim_{\Delta x \to 0}} -\dfrac{1}{x(x + \Delta x)} \\[5ex] \dfrac{dy}{dx} = -\dfrac{1}{x(x + 0)} \\[5ex] \dfrac{dy}{dx} = -\dfrac{1}{x(x)} \\[5ex] \dfrac{dy}{dx} = -\dfrac{1}{x^2} \\[5ex] \therefore \dfrac{dy}{dx} = -\dfrac{1}{x^2} \\[7ex] \underline{Second\:\:Method:\:\:By\:\:Formula} \\[3ex] f'(x) = \displaystyle{\lim_{h \to 0}}\: \dfrac{f(x + h) - f(x)}{h} \\[5ex] f(x) = \dfrac{1}{x} \\[5ex] f(x + h) = \dfrac{1}{x + h} \\[5ex] \underline{Numerator} \\[3ex] f(x + h) - f(x) \\[3ex] = \dfrac{1}{x + h} - \dfrac{1}{x} \\[5ex] = \dfrac{x - (x + h)}{x(x + h)} \\[5ex] = \dfrac{x - x - h}{x(x + h)} \\[5ex] = \dfrac{-h}{x(x + h)} \\[5ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{-h}{x(x + h)} \div h \\[5ex] = \dfrac{-h}{x(x + h)} * \dfrac{1}{h} \\[5ex] = \dfrac{-1}{x(x + h)} \\[5ex] f'(x) = \displaystyle{\lim_{h \to 0}}\: \dfrac{-1}{x(x + h)} \\[5ex] f'(x) = \dfrac{-1}{x(x + 0)} \\[5ex] f'(x) = \dfrac{-1}{x(x)} \\[5ex] f'(x) = \dfrac{-1}{x^2} \\[5ex] \therefore \dfrac{dy}{dx} = -\dfrac{1}{x^2} $

$ y = \dfrac{1}{x} = x^{-1} \\[5ex] \dfrac{dy}{dx} = -1 * x^{-1 - 1} \\[5ex] \dfrac{dy}{dx} = -x^{-2} \\[5ex] \dfrac{dy}{dx} = -\dfrac{1}{x^2} $

We can solve this question in two ways.
Use any method you prefer.

$ \underline{First\:\:Method:\:\:By\:\:Definition} \\[3ex] y = \sqrt{x}...eqn.(1) \\[3ex] y + \Delta y = \sqrt{x + \Delta x} \\[3ex] Subtract\:\:eqn.(1) \\[3ex] y + \Delta y - y = \sqrt{x + \Delta x} - \sqrt{x} \\[3ex] \Delta y = \sqrt{x + \Delta x} - \sqrt{x} \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:\Delta x \\[3ex] \dfrac{\Delta y}{\Delta x} = \dfrac{\sqrt{x + \Delta x} - \sqrt{x}}{\Delta x} \\[5ex] Simplify\:\:the\:\:RHS \\[3ex] Apply\:\:Difference\:\:of\:\:Two\:\:Squares \\[3ex] $ In that case, we need to multiply the numerator and the denominator by the conjugate of the numerator

$ Numerator = \sqrt{x + \Delta x} - \sqrt{x} \\[3ex] Conjugate = \sqrt{x + \Delta x} + \sqrt{x} \\[3ex] $ That would give us a real number.
We do this so we can simplify the numerator and keep moving, else we hit a snag.

$ RHS = \dfrac{\sqrt{x + \Delta x} - \sqrt{x}}{\Delta x} \\[5ex] Simplify \\[3ex] = \dfrac{\sqrt{x + \Delta x} - \sqrt{x}}{\Delta x} * \dfrac{\sqrt{x + \Delta x} + \sqrt{x}}{\sqrt{x + \Delta x} + \sqrt{x}} \\[5ex] \\[5ex] = \dfrac{(\sqrt{x + \Delta x} - \sqrt{x})(\sqrt{x + \Delta x} + \sqrt{x})}{\Delta x(\sqrt{x + \Delta x} + \sqrt{x})} \\[5ex] (\sqrt{x + \Delta x} - \sqrt{x})(\sqrt{x + \Delta x} + \sqrt{x}) = (\sqrt{x + \Delta x})^2 - (\sqrt{x})^2...Difference\:\:of\:\:Two\:\:Squares \\[3ex] (\sqrt{x + \Delta x})^2 - (\sqrt{x})^2 = x + \Delta x - x = \Delta x \\[3ex] = \dfrac{\Delta x}{\Delta x(\sqrt{x + \Delta x} + \sqrt{x})} \\[5ex] = \dfrac{1}{\sqrt{x + \Delta x} + \sqrt{x}} \\[5ex] \rightarrow \dfrac{\Delta y}{\Delta x} = \dfrac{1}{\sqrt{x + \Delta x} + \sqrt{x}} \\[5ex] Limit\:\:as\:\:\Delta x\rightarrow 0 \\[3ex] \displaystyle{\lim_{\Delta x \to 0}} \dfrac{\Delta y}{\Delta x} = \displaystyle{\lim_{\Delta x \to 0}} \dfrac{1}{\sqrt{x + \Delta x} + \sqrt{x}} \\[5ex] \dfrac{dy}{dx} = \dfrac{1}{\sqrt{x + 0} + \sqrt{x}} \\[5ex] \dfrac{dy}{dx} = \dfrac{1}{\sqrt{x} + \sqrt{x}} \\[5ex] \dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}} \\[5ex] \therefore \dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}} \\[7ex] \underline{Second\:\:Method:\:\:By\:\:Formula} \\[3ex] f'(x) = \displaystyle{\lim_{h \to 0}}\: \dfrac{f(x + h) - f(x)}{h} \\[5ex] f(x) = \sqrt{x} \\[3ex] f(x + h) = \sqrt{x + h} \\[3ex] \underline{Numerator} \\[3ex] f(x + h) - f(x) \\[3ex] = \sqrt{x + h} - \sqrt{x} \\[3ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{\sqrt{x + h} - \sqrt{x}}{h} \\[5ex] $ We need to multiply the numerator and the denominator by the conjugate of the numerator so we can simplify the function.
Conjugate of the numerator = $\sqrt{x + h} + \sqrt{x}$

$ = \dfrac{\sqrt{x + h} - \sqrt{x}}{h} * \dfrac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}} \\[5ex] = \dfrac{(\sqrt{x + h})^2 - (\sqrt{x})^2}{h(\sqrt{x + h} + \sqrt{x})} \\[5ex] = \dfrac{(x + h) - x}{h(\sqrt{x + h} + \sqrt{x})} \\[5ex] = \dfrac{x + h - x}{h(\sqrt{x + h} + \sqrt{x})} \\[5ex] = \dfrac{h}{h(\sqrt{x + h} + \sqrt{x})} \\[5ex] = \dfrac{1}{\sqrt{x + h} + \sqrt{x}} \\[5ex] f'(x) = \displaystyle{\lim_{h \to 0}}\: \dfrac{1}{\sqrt{x + h} + \sqrt{x}} \\[5ex] f'(x) = \dfrac{1}{\sqrt{x + 0} + \sqrt{x}} \\[5ex] f'(x) = \dfrac{1}{\sqrt{x} + \sqrt{x}} \\[5ex] f'(x) = \dfrac{1}{2\sqrt{x}} \\[5ex] \therefore \dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}} $

$ y = \sqrt{x} \\[3ex] \sqrt{x} = x^{\dfrac{1}{2}} ...Law\:\:7...Exp \\[5ex] \rightarrow y = x^{\dfrac{1}{2}} \\[5ex] \dfrac{dy}{dx} = \dfrac{1}{2} * x^{\dfrac{1}{2} - 1} \\[5ex] \dfrac{dy}{dx} = \dfrac{1}{2} * x^{\dfrac{1}{2} - \dfrac{2}{2}} \\[5ex] \dfrac{dy}{dx} = \dfrac{1}{2} * x^-{\dfrac{1}{2}} \\[5ex] x^-{\dfrac{1}{2}} = \dfrac{1}{x^{\dfrac{1}{2}}} ...Law\:\:6...Exp \\[7ex] \dfrac{dy}{dx} = \dfrac{1}{2} * \dfrac{1}{x^{\dfrac{1}{2}}} \\[7ex] \dfrac{dy}{dx} = \dfrac{1}{2} * \dfrac{1}{\sqrt{x}} \\[7ex] \therefore \dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}} $

Product Rule, Sum/Difference Rule, Power Rule

$ y = (2x + 5)^2(x - 4) \\[3ex] u = (2x + 5)^2 = (2x + 5)(2x + 5) = 4x^2 + 10x + 10x + 25 = 4x^2 + 20x + 25 \\[3ex] \dfrac{du}{dx} = 8x + 20 \\[5ex] v = x - 4 \\[3ex] \dfrac{dv}{dx} = 1 \\[5ex] \dfrac{dy}{dx} = u\dfrac{dv}{dx} + v\dfrac{du}{dx}...Product\:\:Rule \\[5ex] \dfrac{dy}{dx} = (4x^2 + 20x + 25)(1) + (x - 4)(8x + 20) \\[5ex] \dfrac{dy}{dx} = 4x^2 + 20x + 25 + 8x^2 + 20x - 32x - 80 \\[5ex] \dfrac{dy}{dx} = 12x^2 + 8x - 55 \\[3ex] $ This is JAMB exam. You need to solve this within a minute without a calculator.
A quick review at the options shows that option $D.$ is the answer

$ \dfrac{dy}{dx} = (2x + 5)(6x - 11) $

Product Rule, Power Rule

$ y = x\sin x \\[3ex] u = x \\[3ex] u' = 1 \\[3ex] v = \sin x \\[3ex] v' = \cos x \\[3ex] y' = uv' + vu' ...Product\:\:Rule \\[3ex] y' = x(\cos x) + \sin x(1) \\[3ex] y' = x\cos x + \sin x \\[3ex] x = \dfrac{\pi}{2} = \dfrac{180^\circ}{2} = 90^\circ \\[5ex] y' = \dfrac{\pi}{2}\cos90^\circ + \sin90^\circ \\[5ex] \cos 90^\circ = 0 \\[3ex] \sin 90^\circ = 1 \\[3ex] \rightarrow y' = \dfrac{\pi}{2} * 0 + 1 \\[5ex] y' = 0 + 1 \\[3ex] y' = 1 $

Power Rule

$ y = x^2 - \dfrac{1}{x} \\[5ex] \dfrac{1}{x} = x^{-1}...Law\:\:6...Exp \\[3ex] y = x^2 - x^{-1} \\[3ex] \dfrac{dy}{dx} = 2 * x^{2 - 1} - -1 * x^{-1 - 1} \\[5ex] \dfrac{dy}{dx} = 2 * x^1 + 1 * x^{-2} \\[5ex] \dfrac{dy}{dx} = 2x + x^{-2} \\[5ex] \dfrac{dy}{dx} = 2x + \dfrac{1}{x^2} $

The slope is the derivative
Power Rule

$ y = 2x^2 + 5x - 3 \\[3ex] y' = 4x + 5 \\[3ex] (1, 4) \rightarrow x = 1, y = 4 \\[3ex] y' = 4(1) + 5 \\[3ex] y' = 4 + 5 \\[3ex] y' = 9 $

The maximum value is the $f(\theta)-value$
The derivative is used to find the maximum value.
First, find the derivative first. Second, set it to zero and solve for $\theta$. Third, substitute the value of $\theta$ in the main equation in order to find the $f(\theta)$
Quotient Rule, Product Rule, Sum Rule, Power Rule

$ f(\theta) = \dfrac{4}{6 + 2\cos\theta} \\[5ex] u = 4 \\[3ex] \dfrac{du}{d\theta} = 0 \\[5ex] v = 6 + 2\cos\theta \\[3ex] \dfrac{dv}{d\theta} = 2 * -\sin\theta + \cos\theta * 0 = -2\sin\theta + 0 = -2\sin\theta \\[5ex] f'(\theta) = \dfrac{v\dfrac{du}{d\theta} - u\dfrac{du}{d\theta}}{v^2} \\[5ex] f'(\theta) = \dfrac{(6 + 2\cos\theta)(0) - 4(-2\sin\theta)}{(6 + 2\cos\theta)^2} \\[5ex] f'(\theta) = \dfrac{-4(-2\sin\theta)}{(6 + 2\cos\theta)^2} \\[5ex] Set\:\:f'(\theta) = 0 \\[3ex] \dfrac{-4(-2\sin\theta)}{(6 + 2\cos\theta)^2} = 0 \\[5ex] -4(-2\sin\theta) = 0((6 + 2\cos\theta)^2) \\[3ex] 8\sin\theta = 0 \\[3ex] \sin\theta = \dfrac{0}{8} \\[5ex] \sin\theta = 0 \\[3ex] \theta = 0, \pi, 2\pi ...Unit\:\:Circle\:\:Trigonometry \\[3ex] But\:\:\dfrac{\pi}{2} \lt \theta \lt 2\pi...Question \\[3ex] \therefore \theta = \pi = 180^\circ \\[3ex] Substitute\:\:for\:\:\theta\:\:in\:\:the\:\:main\:\:equation \\[3ex] f(\theta) = \dfrac{4}{6 + 2\cos180^\circ} \\[5ex] \cos180^\circ = -1 \\[3ex] f(\theta) = \dfrac{4}{6 + 2(-1)} \\[5ex] f(\theta) = \dfrac{4}{6 - 2} \\[5ex] f(\theta) = \dfrac{4}{4} \\[5ex] f(\theta) = 1 $

The maximum value is the $y-value$
The derivative is used to find the maximum value.
We shall use it as our first method.
First, find the derivative first. Second, set it to zero and solve for $\theta$. Third, substitute the value of $\theta$ in the main equation in order to find the $f(\theta)$
Power Rule


$ y = 3x^2 - x^3 \\[3ex] y' = 6x - 3x^2 \\[3ex] Set\:\:y = 0 \\[3ex] 6x - 3x^2 = 0 \\[3ex] 3x(2 - x) = 0 \\[3ex] 3x = 0 \:\:OR\:\: 2 - x = 0 \\[3ex] x = \dfrac{0}{3} \:\:OR\:\: 2 - 0 = x \\[5ex] x = 0 \:\:OR\:\: x = 2 \\[3ex] $ To find the maximum value of $y$ in this case (we have two values of $x$), we can solve it two ways.
Use whichever way you prefer.

$ \underline{First\:\:Method - Second\:\:Derivative\:\:Test} \\[3ex] y' = 6x - 3x^2 \\[3ex] y'' = 6 - 6x = 6(1 - x) \\[3ex] For\:\:x = 0 \\[3ex] y'' = 6(1 - 0) = 6(1) = 6 \\[3ex] 6 \gt 0 \\[3ex] x = 0 \:\:gives\:\:minimum\:\:y-value \\[3ex] For\:\:x = 2 \\[3ex] y'' = 6(1 - 2) = 6(-1) = -6 \\[3ex] -6 \lt 0 \\[3ex] x = 2 \:\:gives\:\:maximum\:\:y-value \\[3ex] Substitute\:\:x = 2\:\:in\:\:the\:\:main\:\:equation \\[3ex] y = 3(2)^2 - 2^3 \\[3ex] y = 3(4) - 8 \\[3ex] y = 12 - 8 \\[3ex] y = 4 \\[5ex] \underline{Second\:\:Method - Test\:\:Values} \\[3ex] For\:\:x = 0 \\[3ex] y = 3(0)^2 - 0^3 \\[3ex] y = 3(0) - 0 \\[3ex] y = 0 - 0 \\[3ex] y = 0 \\[3ex] For\:\:x = 2 \\[3ex] y = 3(2)^2 - 2^3 \\[3ex] y = 3(4) - 8 \\[3ex] y = 12 - 8 \\[3ex] y = 4 \\[5ex] 4 \gt 0 \\[3ex] \therefore y = 4 $

$ y = x - 2x^3...eqn.(1) \\[3ex] y + \Delta y = (x + \Delta x) - [2(x + \Delta x)^3] \\[3ex] (x + \Delta x)^3 = x^3 + 3x^2\Delta x + 3x\Delta^2x + \Delta^3x \\[3ex] 2(x + \Delta x)^3 = 2x^3 + 6x^2\Delta x + 6x\Delta^2x + 2\Delta^3x \\[3ex] (x + \Delta x) - [2(x + \Delta x)^3] = (x + \Delta x) - (2x^3 + 6x^2\Delta x + 6x\Delta^2x + 2\Delta^3x) \\[3ex] (x + \Delta x) - [2(x + \Delta x)^3] = x + \Delta x - 2x^3 - 6x^2\Delta x - 6x\Delta^2x - 2\Delta^3x \\[3ex] \rightarrow y + \Delta y = x + \Delta x - 2x^3 - 6x^2\Delta x - 6x\Delta^2x - 2\Delta^3x \\[3ex] Subtract\:\:eqn.(1) \\[3ex] y + \Delta y - y = x + \Delta x - 2x^3 - 6x^2\Delta x - 6x\Delta^2x - 2\Delta^3x - (x - 2x^3) \\[3ex] \Delta y = x + \Delta x - 2x^3 - 6x^2\Delta x - 6x\Delta^2x - 2\Delta^3x - x + 2x^3 \\[3ex] \Delta y = \Delta x - 6x^2\Delta x - 6x\Delta^2x - 2\Delta^3x \\[3ex] Divide\:\:both\:\:sides\:\:by\:\:\Delta x \\[3ex] \dfrac{\Delta y}{\Delta x} = \dfrac{\Delta x - 6x^2\Delta x - 6x\Delta^2x - 2\Delta^3x }{\Delta x} \\[5ex] \dfrac{\Delta y}{\Delta x} = \dfrac{\Delta x}{\Delta x} - \dfrac{6x^2\Delta x}{\Delta x} - \dfrac{6x\Delta^2x}{\Delta x} - \dfrac{2\Delta^3x}{\Delta x} \\[5ex] \dfrac{\Delta y}{\Delta x} = 1 - 6x^2 - 6x\Delta x - 2\Delta^2x \\[5ex] Limit\:\:as\:\:\Delta x\rightarrow 0 \\[3ex] \displaystyle{\lim_{\Delta x \to 0}} \dfrac{\Delta y}{\Delta x} = \displaystyle{\lim_{\Delta x \to 0}} (1 - 6x^2 - 6x\Delta x - 2\Delta^2x) \\[5ex] \dfrac{dy}{dx} = 1 - 6x^2 - 6x(0) - 2(0)^2 \\[5ex] \therefore \dfrac{dy}{dx} = 1 - 6x^2 $

Power Rule

$ y = x - 2x^3 \\[3ex] y = x^1 - 2x^3 \\[3ex] \dfrac{dy}{dx} = (1 * x^{1 - 1}) - (3 * 2 * x^{3 - 1}) \\[5ex] \dfrac{dy}{dx} = (1 * x^0) - (6 * x^2) \\[5ex] x^0 = 1 ...Law\:\:3...Exp \\[3ex] \dfrac{dy}{dx} = (1 * 1) - (6x^2) \\[5ex] \dfrac{dy}{dx} = 1 - 6x^2 $

$ (17.1) \\[3ex] y = 3x^3 + 6x^2 + x - 4 \\[3ex] \dfrac{dy}{dx} = 9x^2 + 12x + 1 \\[5ex] (17.2) \\[3ex] yx - y = 2x^2 - 2x \\[3ex] Factor\:\:by\:\:GCF \\[3ex] y(x - 1) = 2x(x - 1) \\[3ex] y = \dfrac{2x(x - 1)}{x - 1} \\[5ex] y = 2x \\[3ex] \dfrac{dy}{dx} = 2 $

Quotient Rule

$ f(x) = \tan x \\[3ex] = \dfrac{\sin x}{\cos x} ...Quotient\;\;Identity \\[5ex] For\;\;f(x) = \dfrac{u}{v} \\[5ex] u = \sin x \\[3ex] u' = \cos x \\[3ex] v = \cos x \\[3ex] v' = -\sin x \\[3ex] f'(x) = \dfrac{vu' - uv'}{v^2} \\[5ex] \underline{Numerator} \\[3ex] vu' - uv' \\[3ex] = (\cos x)(\cos x) - (\sin x)(-\sin x) \\[3ex] = \cos^2 x + \sin^2 x \\[3ex] = 1...Pythagorean\;\;Identity \\[3ex] \underline{Denominator} \\[3ex] v^2 = \cos^2 x \\[3ex] \implies \\[3ex] f'(x) = \dfrac{1}{\cos^2 x} \\[5ex] \therefore \dfrac{d}{dx} \tan(x) = \dfrac{1}{\cos^2(x)} $

Chain Rule, Standard Derivatives

$ y = \ln|\ln|\ln x|| \\[3ex] Let: \\[3ex] p = \ln x \:\:\:\: \dfrac{dp}{dx} = \dfrac{1}{x} \\[5ex] q = |p| \:\:\:\: \dfrac{dq}{dp} = \dfrac{|p|}{p} \\[5ex] r = \ln q \:\:\:\: \dfrac{dr}{dq} = \dfrac{1}{q} \\[5ex] s = |r| \:\:\:\: \dfrac{ds}{dr} = \dfrac{|r|}{r} \\[5ex] y = \ln s \:\:\:\: \dfrac{dy}{ds} = \dfrac{1}{s} \\[5ex] \dfrac{dy}{dx} = \dfrac{dp}{dx} * \dfrac{dq}{dp} * \dfrac{dr}{dq} * \dfrac{ds}{dr} * \dfrac{dy}{ds} \\[5ex] \dfrac{dy}{dx} = \dfrac{1}{x} * \dfrac{|p|}{p} * \dfrac{1}{q} * \dfrac{|r|}{r} * \dfrac{1}{s} \\[5ex] Substitute\:\:back \\[3ex] \dfrac{dy}{dx} = \dfrac{1}{x} * \dfrac{|\ln x|}{\ln x} * \dfrac{1}{|\ln x|} * \dfrac{|\ln q|}{\ln q} * \dfrac{1}{|\ln q|} \\[5ex] \dfrac{dy}{dx} = \dfrac{1}{x} * \dfrac{1}{\ln x} * \dfrac{1}{\ln q} \\[5ex] \dfrac{dy}{dx} = \dfrac{1}{x} * \dfrac{1}{\ln x} * \dfrac{1}{\ln |p|} \\[5ex] \dfrac{dy}{dx} = \dfrac{1}{x} * \dfrac{1}{\ln x} * \dfrac{1}{\ln |\ln x|} \\[5ex] \dfrac{dy}{dx} = \dfrac{1 * 1 * 1}{x * \ln x * \ln|\ln x|} \\[5ex] \dfrac{dy}{dx} = \dfrac{1}{x\ln x\ln|\ln x|} $

Implicit Differentiation, Product Rule, Power Rule

$ x^6y + y^6x = 9 \\[3ex] \dfrac{d(x^6y)}{dx} = x^6 * \dfrac{dy}{dx} + y * 6x^5 ...Product\:\:Rule \\[5ex] \dfrac{d(x^6y)}{dx} = x^6\dfrac{dy}{dx} + 6x^5y \\[5ex] \dfrac{d(y^6x)}{dx} = y^6 * 1 + x * 6y^5\dfrac{dy}{dx} ...Product\:\:Rule \\[5ex] \dfrac{d(y^6x)}{dx} = y^6 + 6xy^5\dfrac{dy}{dx} \\[5ex] \dfrac{d(9)}{dx} = 0 \\[5ex] \rightarrow x^6\dfrac{dy}{dx} + 6x^5y + y^6 + 6xy^5\dfrac{dy}{dx} = 0 \\[5ex] x^6\dfrac{dy}{dx} + 6xy^5\dfrac{dy}{dx} = 0 - 6x^5y - y^6 \\[5ex] \dfrac{dy}{dx}(x^6 + 6xy^5) = -6x^5y - y^6 \\[5ex] \dfrac{dy}{dx} = \dfrac{-6x^5y - y^6}{x^6 + 6xy^5} \\[5ex] \dfrac{dy}{dx} = -\dfrac{(6x^5y + y^6)}{x^6 + 6xy^5} $

We can solve this question in two ways.
Use any method you prefer.

$ \underline{First\:\:Method:\:\:By\:\:Definition} \\[3ex] f(x) = 3x^2 + 4x - 1 \\[3ex] f(x + \Delta x) = 3(x + \Delta x)^2 + 4(x + \Delta x) - 1 \\[3ex] = 3[(x + \Delta x)(x + \Delta x)] + 4x + 4\Delta x - 1 \\[3ex] = 3[x^2 + x\Delta x + x\Delta x + (\Delta x)^2] + 4x + 4\Delta x - 1 \\[3ex] = 3[x^2 + 2x\Delta x + (\Delta x)^2] + 4x + 4\Delta x - 1 \\[3ex] = 3x^2 + 6x\Delta x + 3(\Delta x)^2 + 4x + 4\Delta x - 1 \\[3ex] = 3x^2 + 4x - 1 + 6x\Delta x + 4\Delta x + 3(\Delta x)^2 \\[5ex] f(x + \Delta x) - f(x) \\[3ex] = 3x^2 + 4x - 1 + 6x\Delta x + 4\Delta x + 3(\Delta x)^2 - (3x^2 + 4x - 1) \\[3ex] = 3x^2 + 4x - 1 + 6x\Delta x + 4\Delta x + 3(\Delta x)^2 - 3x^2 - 4x + 1 \\[3ex] = 6x\Delta x + 4\Delta x + 3(\Delta x)^2 \\[3ex] = \Delta x(6x + 4 + 3\Delta x) \\[5ex] \dfrac{f(x + \Delta x) - f(x)}{\Delta x} \\[5ex] = \dfrac{\Delta x(6x + 4 + 3\Delta x)}{\Delta x} \\[5ex] = 6x + 4 + 3\Delta x \\[5ex] f'(x) = \displaystyle{\lim_{\Delta x \to 0}}\:\: 6x + 4 + 3\Delta x \\[3ex] f'(x) = 6x + 4 + 3(0) \\[3ex] f'(x) = 6x + 4 \\[5ex] \underline{Second\:\:Method:\:\:By\:\:Formula} \\[3ex] f'(x) = \displaystyle{\lim_{h \to 0}}\: \dfrac{f(x + h) - f(x)}{h} \\[5ex] f(x) = 3x^2 + 4x - 1 \\[3ex] f(x + h) = 3(x + h)^2 + 4(x + h) - 1 \\[3ex] = 3[(x + h)(x + h)] + 4x + 4h - 1 \\[3ex] = 3(x^2 + hx + hx + h^2) + 4x + 4h - 1 \\[3ex] = 3(x^2 + 2hx + h^2) + 4x + 4h - 1 \\[3ex] = 3x^2 + 6hx + 3h^2 + 4x + 4h - 1 \\[3ex] \underline{Numerator} \\[3ex] f(x + h) - f(x) \\[3ex] = 3x^2 + 6hx + 3h^2 + 4x + 4h - 1 - (3x^2 + 4x - 1) \\[3ex] = 3x^2 + 6hx + 3h^2 + 4x + 4h - 1 - 3x^2 - 4x + 1 \\[3ex] = 6hx + 3h^2 + 4h \\[3ex] = h(6x + 3h + 4) \\[3ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{h(6x + 3h + 4)}{h} \\[5ex] = 6x + 3h + 4 \\[3ex] f'(x) = \displaystyle{\lim_{h \to 0}}\: 6x + 3h + 4 \\[3ex] f'(x) = 6x + 3(0) + 4 \\[3ex] f'(x) = 6x + 4 $

Function of a Function Rule, Power Rule

$ y = \left(3x - x^2\right)^5 \\[3ex] Let\;\; p = 3x - x^2 \;\;\; means\;\;that\;\; y = p^5 \\[3ex] \dfrac{dp}{dx} = 3 - 2x \\[5ex] \dfrac{dy}{dp} = 5p^4 \\[3ex] \dfrac{dy}{dx} = \dfrac{dy}{dp} * \dfrac{dp}{dx} ...Chain\;\;Rule \\[5ex] = 5p^4(3 - 2x) \\[3ex] = 5(3x - x^2)^4(3 - 2x) \\[3ex] = 5(3 - 2x)(3x - x^2)^4 $

We can solve this question in two ways.
Use any method you prefer.

$ \underline{First\:\:Method:\:\:By\:\:Definition} \\[3ex] f(x) = 3x^2 + 2x - 1 \\[3ex] f(x + \Delta x) = 3(x + \Delta x)^2 + 2(x + \Delta x) - 1 \\[3ex] = 3[(x + \Delta x)(x + \Delta x)] + 2x + 2\Delta x - 1 \\[3ex] = 3[x^2 + x\Delta x + x\Delta x + (\Delta x)^2] + 2x + 2\Delta x - 1 \\[3ex] = 3[x^2 + 2x\Delta x + (\Delta x)^2] + 2x + 2\Delta x - 1 \\[3ex] = 3x^2 + 6x\Delta x + 3(\Delta x)^2 + 2x + 2\Delta x - 1 \\[3ex] = 3x^2 + 2x - 1 + 6x\Delta x + 2\Delta x + 3(\Delta x)^2 \\[5ex] f(x + \Delta x) - f(x) \\[3ex] = 3x^2 + 2x - 1 + 6x\Delta x + 2\Delta x + 3(\Delta x)^2 - (3x^2 + 2x - 1) \\[3ex] = 3x^2 + 2x - 1 + 6x\Delta x + 2\Delta x + 3(\Delta x)^2 - 3x^2 - 2x + 1 \\[3ex] = 6x\Delta x + 2\Delta x + 3(\Delta x)^2 \\[3ex] = \Delta x(6x + 2 + 3\Delta x) \\[5ex] \dfrac{f(x + \Delta x) - f(x)}{\Delta x} \\[5ex] = \dfrac{\Delta x(6x + 2 + 3\Delta x)}{\Delta x} \\[5ex] = 6x + 2 + 3\Delta x \\[5ex] f'(x) = \displaystyle{\lim_{\Delta x \to 0}}\:\: 6x + 2 + 3\Delta x \\[3ex] f'(x) = 6x + 2 + 3(0) \\[3ex] f'(x) = 6x + 2 \\[5ex] \underline{Second\:\:Method:\:\:By\:\:Formula} \\[3ex] f'(x) = \displaystyle{\lim_{h \to 0}}\: \dfrac{f(x + h) - f(x)}{h} \\[5ex] f(x) = 3x^2 + 2x - 1 \\[3ex] f(x + h) = 3(x + h)^2 + 2(x + h) - 1 \\[3ex] = 3[(x + h)(x + h)] + 2x + 2h - 1 \\[3ex] = 3(x^2 + hx + hx + h^2) + 2x + 2h - 1 \\[3ex] = 3(x^2 + 2hx + h^2) + 2x + 2h - 1 \\[3ex] = 3x^2 + 6hx + 3h^2 + 2x + 2h - 1 \\[3ex] \underline{Numerator} \\[3ex] f(x + h) - f(x) \\[3ex] = 3x^2 + 6hx + 3h^2 + 2x + 2h - 1 - (3x^2 + 2x - 1) \\[3ex] = 3x^2 + 6hx + 3h^2 + 2x + 2h - 1 - 3x^2 - 2x + 1 \\[3ex] = 6hx + 3h^2 + 2h \\[3ex] = h(6x + 3h + 2) \\[3ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{h(6x + 3h + 2)}{h} \\[5ex] = 6x + 3h + 2 \\[3ex] f'(x) = \displaystyle{\lim_{h \to 0}}\: 6x + 3h + 2 \\[3ex] f'(x) = 6x + 3(0) + 2 \\[3ex] f'(x) = 6x + 2 $

Quotient Rule, Power Rule

$ f(x) = \dfrac{\tan x}{x^3} \\[5ex] For\;\;f(x) = \dfrac{u}{v} \\[5ex] u = \tan x \\[3ex] u' = \sec^2 x \\[3ex] v = x^3 \\[3ex] v' = 3x^2 \\[3ex] f'(x) = \dfrac{vu' - uv'}{v^2} \\[5ex] \underline{Numerator} \\[3ex] vu' - uv' \\[3ex] = (x^3)(\sec^2 x) - (\tan x)(3x^2) \\[3ex] = x^3\sec^2 x - 3x^2\tan x \\[3ex] \underline{Denominator} \\[3ex] v^2 = (x^3)^2 = x^6 \\[3ex] \implies \\[3ex] f'(x) \\[3ex] = \dfrac{x^3\sec^2 x - 3x^2\tan x}{x^6} \\[5ex] = \dfrac{x^2(x\sec^2 x - 3\tan x)}{x^6} \\[5ex] = \dfrac{x\sec^2 x - 3\tan x}{x^4} $

$ (25.1) \\[3ex] Function:\;\;h(x) = ax^3 + bx^2 \\[3ex] For\;\;Point\;B(4, 32) \\[3ex] x = 4 \\[3ex] y = h(x) = 32 \\[3ex] \implies \\[3ex] 32 = a(4)^3 + b(4)^2 \\[3ex] 32 = 64a + 16b \\[3ex] 64a + 16b = 32 \\[3ex] \implies \\[3ex] 4a + b = 2 ...eqn.(1) \\[3ex] Also: \\[3ex] h(x) = ax^3 + bx^2 \\[3ex] h'(x) = 3ax^2 + 2bx \\[3ex] At\;\;turning\;\;point,\;\;h'(x) = 0 \\[3ex] For\;\;Turning\;\;Point\;B(4, 32) \\[3ex] x = 4 \\[3ex] \implies \\[3ex] h'(4) = 0 \\[3ex] h'(4) = 3a(4)^2 + 2b(4) = 0 \\[3ex] 48a + 8b = 0 \\[3ex] \implies \\[3ex] 6a + b = 0 ...eqn.(2) \\[3ex] eqn.(2) - eqn.(1) \\[3ex] 6a - 4a = 0 - 2 \\[3ex] 2a = -2 \\[3ex] a = -\dfrac{2}{2} \\[5ex] a = -1 \\[3ex] Substitute\;\;a = -1 \;\;into\;\;eqn.(1) \\[3ex] 4(-1) + b = 2 \\[3ex] -4 + b = 2 \\[3ex] b = 2 + 4 \\[3ex] b = 6 \\[3ex] \therefore h(x) = -x^3 + 6x^2 \\[3ex] (25.2) \\[3ex] A = x-intercept\;\;of\;\;h(x) \\[3ex] \implies \\[3ex] h(x) = 0 \\[3ex] -x^3 + 6x^2 = 0 \\[3ex] x^2(-x + 6) = 0 \\[3ex] x^2 = 0 \;\;\;OR\;\;\; -x + 6 = 0 \\[3ex] x = \pm \sqrt{0} \;\;\;OR\;\;\; 6 = 0 + x \\[3ex] x = \pm 0 \;\;\;OR\;\;\; 6 = x \\[3ex] Based\;\;on\;\;the\;\;graph:\;\; x \ne \pm 0 \\[3ex] x = 6 \\[3ex] \therefore coordinates\;\;of\;\;A = (6, 0) \\[3ex] (25.3) \\[3ex] (25.3.1) \\[3ex] Based\;\;on\;\;the\;\;Graph \\[3ex] h(x)\uparrow for\;\; x \in (0, 4) \\[3ex] (25.3.2) \\[3ex] Based\;\;on\;\;the\;\;function \\[3ex] h(x) = -x^3 + 6x^2 \\[3ex] h'(x) = -3x^2 + 12x \\[3ex] h''(x) = -6x + 12 \\[3ex] h''(x) = 0 \\[3ex] \implies \\[3ex] -6x + 12 = 0 \\[3ex] -6x = -12 \\[3ex] x = \dfrac{-12}{-6} \\[5ex] Inflection\;\;Point:\;\;x = 2 \\[3ex] Intervals:\;\; x \lt 2 \;\;and\;\; x \gt 2 \\[3ex] $

$h''(x) = -6x + 12 $

Interval	$x \lt 2$	$x \gt 2$
Test Value	$x = 1$	$x = 3$
Sign Test	$$ -6(1) + 12 \\[3ex] -6 + 12 = 6 \\[3ex] 6 \gt 0 \\[3ex] positive $$	$$ -6(3) + 12 \\[3ex] -18 + 12 = -6 \\[3ex] -6 \lt 0 \\[3ex] negative $$
Conclusion	Concave up	Concave down

$ h(x)\frown for\;\; x \gt 2 \\[3ex] h(x)\frown for\;\; x \in (2, \infty) \\[3ex] (25.4) \\[3ex] -(x - 1)^3 + 6(x - 1)^2 - k = 0 \\[3ex] -(x - 1)^3 + 6(x - 1)^2 = k \\[3ex] k = -(x - 1)^3 + 6(x - 1)^2 \\[3ex] h(x) = -x^3 + 6x^2 \\[3ex] h(x - 1) = -(x - 1)^3 + 6(x - 1)^2 \\[3ex] \implies k = h(x - 1) \\[3ex] k = graph\;\;of\;\;h(x - 1) \\[3ex] 1st\;\;find\;\;the\;\;y-intercept...value\;\;of\;\;k\;\;when\;\;x = 0 \\[3ex] k = -(0 - 1)^3 + 6(0 - 1)^2 \\[3ex] k = -(-1)^3 + 6(-1)^2 \\[3ex] k = -(-1) + 6(1) \\[3ex] k = 1 + 6 \\[3ex] k = 7 \\[3ex] Negative\;\;root \implies x \lt 0 \\[3ex] When\;\;x \lt 0,\;\;k \lt 7 \\[3ex] Let\;\;us\;\;find\;\;the\;\;maximum\;\;value\;\;for\;\;the\;\;positive\;\;roots \\[3ex] h(x - 1) = -(x - 1)^3 + 6(x - 1)^2 \\[3ex] h(x - 1)' = -3(x - 1)^2 + 12(x - 1) \\[3ex] h(x - 1)' = 0 \\[3ex] \implies \\[3ex] -3(x - 1)^2 + 12(x - 1) = 0 \\[3ex] -3(x^2 - 2x + 1) + 12x - 12 = 0 \\[3ex] -3x^2 + 6x - 3 + 12x - 12 = 0 \\[3ex] -3x^2 + 18x - 15 = 0 \\[3ex] x^2 - 6x + 5 = 0 \\[3ex] (x - 5)(x - 1) = 0 \\[3ex] x - 5 = 0 \;\;\;OR\;\;\; x - 1 = 0 \\[3ex] x = 5 \;\;\;OR\;\;\; x = 1 \\[3ex] For\;\;x = 5 \\[3ex] h(x - 1) = -(5 - 1)^3 + 6(5 - 1)^2 \\[3ex] = -4^3 + 6(4^2) \\[3ex] = -64 + 6(16) \\[3ex] = -64 + 96 \\[3ex] = 32 \\[3ex] For\;\;x = 1 \\[3ex] h(x - 1) = -(1 - 1)^3 + 6(0 - 1)^2 \\[3ex] = -0^3 + 6(-1)^2 \\[3ex] = 0 + 6(1) \\[3ex] = 0 + 6 \\[3ex] = 6 \\[3ex] Maximum\;\;value = 32 \\[3ex] \implies \\[3ex] 7 \lt k \lt 32 $

Power Rule

$ y = x^3 + 2x^2 + 3x - 4 \\[3ex] y' = 3x^2 + 4x + 3 $

NOTE: The measure of the angle, $x$ is in radians, RAD

$ 2\sin x = x \\[3ex] 2\sin x - x = 0 \\[3ex] 2\sin x - x = f(x) \\[3ex] f(x) = 2\sin x - x \\[3ex] f'(x) = 2\cos x - 1 \\[3ex] x_{n + 1} = x_n - \dfrac{f(x)}{f'(x)} \\[5ex] \implies \\[3ex] x_{n + 1} = x_n - \dfrac{2\sin x_n - x_n}{2\cos x_n - 1} \\[5ex] $

$n$	$1$	$2$
$x_n$	$2$	$1.900995594$
$\sin x_n$	$0.9092974268$	$0.9459777535$
$2\sin x_n$	$1.818594854$	$1.891955507$
$\color{purple}{2\sin x_n - x_n}$	$-0.1814051463$	$-0.009040087$
$\cos x_n$	$-0.4161468365$	$-0.3242315372$
$2\cos x_n$	$-0.8322936731$	$-0.6484630743$
$\color{purple}{2\cos x_n - 1}$	$-1.832293673$	$-1.648463074$
$\dfrac{2\sin x_n - x_n}{2\cos x_n - 1}$	$0.0990044058$	$0.0054839487$
$\color{darkblue}{x_n - \dfrac{2\sin x_n - x_n}{2\cos x_n - 1}}$	$\color{darkblue}{1.900995594}$	$\color{darkblue}{1.895511645}$

$ x_2 = 1.900995594 \\[3ex] x_3 = 1.895511645 $

$ y = (x^2 - 1)x \\[3ex] y = x(x^2 - 1) \\[3ex] y' = x(2x) + (x^2 - 1)(1)...Product\;\;Rule \\[3ex] y' = 2x^2 + x^2 - 1 \\[3ex] y' = 3x^2 - 1 \\[3ex] At\;\;the\;\;point(1, 0) \\[3ex] x = 1 \\[3ex] y = 0 \\[3ex] m = y' = 3(1)^2 - 1 \\[3ex] m = 3(1) - 1 \\[3ex] m = 3 - 1 \\[3ex] m = 2 \\[3ex] \underline{Equation\;\;of\;\;the\;\;tangent\;\;to\;\;the\;\;curve} \\[3ex] y - y_1 = m(x - x_1)...Point-Slope\;\;Form \\[3ex] x_1 = 1 \\[3ex] y_1 = 0 \\[3ex] m = 2 \\[3ex] \implies \\[3ex] y - 0 = 2(x - 1) \\[3ex] y = 2(x - 1) \\[3ex] y = 2x - 2 \\[3ex] $ None of the above

$ (a.) \\[3ex] f(x) = \dfrac{7}{x} - 7\ln(x) \\[5ex] f(x) = 7x^{-1} - 7\ln x \\[3ex] f'(x) = -7x^-2 - \left[7\left(\dfrac{1}{x}\right) + \ln x(0)\right] \\[5ex] f'(x) = -7x^{-2} - \dfrac{7}{x} \\[5ex] f'(x) = \dfrac{-7}{x^2} - \dfrac{7}{x} \\[5ex] Simplify\;\;f(x)\;\;and\;\;f'(x) \\[3ex] f(x) = \dfrac{7}{x} - 7\ln(x) \\[5ex] f(x) = \dfrac{7 - 7x\ln x}{x} \\[5ex] f'(x) = \dfrac{-7}{x^2} - \dfrac{7}{x} \\[5ex] f'(x) = \dfrac{-7 - 7x}{x^2} \\[5ex] \dfrac{f(x)}{f'(x)} \\[5ex] = f(x) \div f'(x) \\[3ex] = \dfrac{7 - 7x\ln x}{x} \div \dfrac{-7 - 7x}{x^2} \\[5ex] = \dfrac{7 - 7x\ln x}{x} * \dfrac{x^2}{-7 - 7x} \\[5ex] = \dfrac{7(1 - x\ln x)}{x} * \dfrac{x^2}{-7(1 + x)} \\[5ex] = \dfrac{x(1 - x\ln x)}{-(1 + x)} \\[5ex] = \dfrac{x - x^2\ln x}{-(1 + x)} \\[5ex] = -\dfrac{(x - x^2\ln x)}{1 + x} \\[5ex] = \dfrac{-x + x^2\ln x}{x + 1} \\[5ex] = \dfrac{x^2\ln x - x}{x + 1} \\[5ex] x - \dfrac{f(x)}{f'(x)} \\[5ex] = x - \dfrac{x^2\ln x - x}{x + 1} \\[5ex] = \dfrac{x(x + 1) - (x^2\ln x - x)}{x + 1} \\[5ex] = \dfrac{x^2 + x - x^2\ln x + x}{x + 1} \\[5ex] = \dfrac{x^2 + 2x - x^2\ln x}{x + 1} \\[5ex] x_{n + 1} = x_n - \dfrac{f(x)}{f'(x)} \\[5ex] \implies \\[3ex] x_{n + 1} = \dfrac{x_n^2 + 2x_n - x_n^2\ln x_n}{x_n + 1} \\[5ex] x_1 = 1 \\[3ex] \implies \\[3ex] x_2 = \dfrac{x_1^2 + 2x_1 - x_1^2\ln x_1}{x_1 + 1} \\[5ex] x_2 = \dfrac{1^2 + 2(1) - 1^2 * \ln(1)}{1 + 1} \\[5ex] x_2 = \dfrac{1 + 2 - 1 * 0}{2} \\[5ex] x_2 = \dfrac{1 + 2 - 0}{2} \\[5ex] x_2 = \dfrac{3}{2} \\[7ex] x_3 = \dfrac{x_2^2 + 2x_2 - x_2^2\ln x_2}{x_2 + 1} \\[5ex] \underline{Numerator} \\[3ex] x_2^2 + 2x_2 - x_2^2\ln x_2 \\[3ex] = \left(\dfrac{3}{2}\right)^2 + 2 * \dfrac{3}{2} - \left(\dfrac{3}{2}\right)^2 * \ln \left(\dfrac{3}{2}\right) \\[5ex] = \dfrac{9}{4} + 3 - \dfrac{9}{4} * \ln \left(\dfrac{3}{2}\right) \\[5ex] = \dfrac{21}{4} - \dfrac{9}{4} * \ln \left(\dfrac{3}{2}\right) \\[5ex] \underline{Denominator} \\[3ex] x_2 + 1 \\[3ex] = \dfrac{3}{2} + 1 \\[5ex] = \dfrac{3 + 2}{2} \\[5ex] = \dfrac{5}{2} \\[5ex] Numerator \div Denominator \\[3ex] = \dfrac{21}{4} - \dfrac{9}{4} * \ln \left(\dfrac{3}{2}\right) \div \dfrac{5}{2} \\[5ex] = \dfrac{1}{4}\left[21 - 9\ln\left(\dfrac{3}{2}\right)\right] * \dfrac{2}{5} \\[5ex] = \dfrac{1}{10}\left[21 - 9\ln\left(\dfrac{3}{2}\right)\right] \\[5ex] = \dfrac{21}{10} - \dfrac{9}{10}\ln\left(\dfrac{3}{2}\right) \\[5ex] \therefore x_3 = \dfrac{21}{10} - \dfrac{9}{10}\ln\left(\dfrac{3}{2}\right) \\[5ex] (b.) \\[3ex] x_3 = 2.1 - (0.9 * \ln(1.5)) \\[3ex] x_3 = 1.735081403 \\[3ex] $
$ x_4 = 1.762915391 \\[3ex] $
$ x_5 = 1.763222798 \\[3ex] x_6 = 1.763222834 \\[3ex] Because\;\;x_5 \approx x_6...STOP $

$ y = \dfrac{2x}{x^2 + 1} = \dfrac{p}{k} \\[5ex] p = 2x \\[3ex] p' = 2 \\[3ex] k = x^2 + 1 \\[3ex] k' = 2x \\[3ex] y' = \dfrac{kp' - pk'}{k^2} \\[5ex] y' = \dfrac{(x^2 + 1)(2) - 2x(2x)}{(x^2 + 1)^2} \\[5ex] y' = \dfrac{2x^2 + 2 - 4x^2}{(x^2 + 1)^2} \\[5ex] y' = \dfrac{2 - 2x^2}{(x^2 + 1)^2} \\[5ex] \underline{Stationary\;\;Points} \\[3ex] y' = 0 \\[3ex] \implies \\[3ex] \dfrac{2 - 2x^2}{(x^2 + 1)^2} = 0 \\[5ex] 2 - 2x^2 = 0 \\[3ex] 2 = 2x^2 \\[3ex] 2x^2 = 2 \\[3ex] x^2 = \dfrac{2}{2} \\[5ex] x^2 = 1 \\[3ex] x = \pm \sqrt{1} \\[3ex] x = \pm 1 \\[3ex] y = \dfrac{2x}{x^2 + 1} \\[5ex] when\;\; x = 1 \\[3ex] y = \dfrac{2(1)}{1^2 + 1} \\[5ex] y = \dfrac{2}{1 + 1} \\[5ex] y = \dfrac{2}{2} \\[5ex] y = 1 \\[3ex] (x, y) \rightarrow (1, 1) \\[3ex] when\;\; x = -1 \\[3ex] y = \dfrac{2(-1)}{(-1)^2 + 1} \\[5ex] y = \dfrac{-2}{1 + 1} \\[5ex] y = \dfrac{-2}{2} \\[5ex] y = -1 \\[3ex] (x, y) \rightarrow (-1, -1) \\[3ex] $ The stationary points are: (1, 1) and (-1, -1)

$ y = \dfrac{1}{8} \\[5ex] \dfrac{dy}{dx} = 0 \\[3ex] $ The derivative of any constant is zero

Alternative Explanation

$ y = \dfrac{1}{8} \\[5ex] y = \dfrac{1}{8} * 1 \\[5ex] y = \dfrac{1}{8}x^0...Exp...Law\;3 \\[5ex] \dfrac{dy}{dx} = 0 * \dfrac{1}{8}* x^{0 - 1} \\[5ex] \dfrac{dy}{dx} = 0 $

$ y = 3x^2 - x + 1 \\[3ex] y' = 6x - 1 \\[3ex] gradient = y'\;\;@\;\;x = -1 \\[3ex] gradient = 6(-1) - 1 \\[3ex] gradient = -6 - 1 \\[3ex] gradient = -7 $