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# Solved Examples on Differentiation

You may verify your answers as applicable with the: Differential Calculus Calculators
Prerequisite Topics:
Exponents and Logarithms
Factoring
Difference Quotient
Trigonometry

For JAMB and NZQA Students
Calculators are not allowed. So, the questions are solved in a way that does not require a calculator.

For NSC Students
For the Questions:
Any space included in a number indicates a comma used to separate digits...separating multiples of three digits from behind.
Any comma included in a number indicates a decimal point.
For the Solutions:
Decimals are used appropriately rather than commas
Commas are used to separate digits appropriately.

Solve all questions
Show all work
Use at least two methods whenever applicable
Indicate the method(s) you used as applicable

(1.) NSC Determine $f'(x)$ from first principles if it is given $f(x) = x^2 - 5$

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

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

Power Rule and Difference Rule

$f(x) = x^2 - 5 \\[3ex] f'(x) = 2 * x^{2 - 1} - 5x^0 \\[3ex] = 2 * x^1 - 0 * 5 * x^{0 - 1} \\[3ex] = 2 * x - 0 \\[3ex] f'(x) = 2x$
(3.) JAMB Find the maximum value of $y$ in the equation $y = 1 - 2x - 3x^2$

$A.\:\: \dfrac{4}{3} \\[5ex] B.\:\: \dfrac{5}{4} \\[5ex] C.\:\: \dfrac{3}{4} \\[5ex] D.\:\: \dfrac{5}{3} \\[5ex]$

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.

$\underline{First\:\:Method - Derivatives} \\[3ex] y = 1 - 2x - 3x^2 \\[3ex] \dfrac{dy}{dx} = -2 - 6x \\[5ex] Set\:\:\dfrac{dy}{dx} = 0 \\[5ex] -2 - 6x = 0 \\[3ex] -2 = 6x \\[3ex] 6x = -2 \\[3ex] x = -\dfrac{2}{6} \\[5ex] x = -\dfrac{1}{3} \\[5ex] Substitute\:\:for\:\:x\:\:in\:\:the\:\:main\:\:equation \\[3ex] y = 1 - 2\left(-\dfrac{1}{3}\right) - 3\left(-\dfrac{1}{3}\right)^2 \\[5ex] y = 1 + \dfrac{2}{3} - 3\left(-\dfrac{1}{3}\right)\left(-\dfrac{1}{3}\right) \\[5ex] y = \dfrac{3}{3} + \dfrac{2}{3} - \dfrac{1}{3} \\[5ex] y = \dfrac{3 + 2 - 1}{3} \\[5ex] y = \dfrac{5}{3} \\[5ex] \underline{Second\:\:Method - Vertex\:\:Formula} \\[3ex] y = 1 - 2x - 3x^2 \\[3ex] y = -3x^2 - 2x + 1 \\[3ex] y = ax^2 + bx + c \\[3ex] Compare \\[3ex] a = -3 \\[3ex] b = -2 \\[3ex] Vertex = \left(-\dfrac{b}{2a}, f\left(-\dfrac{b}{2a}\right)\right) \\[5ex] x = -\dfrac{b}{2a} = \dfrac{-(-2)}{2(-3)} = \dfrac{2}{-6} = -\dfrac{2}{6} = -\dfrac{1}{3} \\[5ex] Substitute\:\:for\:\:x\:\:in\:\:the\:\:equation \\[3ex] y = 1 - 2\left(-\dfrac{1}{3}\right) - 3\left(-\dfrac{1}{3}\right)^2 \\[5ex] y = 1 + \dfrac{2}{3} - 3\left(-\dfrac{1}{3}\right)\left(-\dfrac{1}{3}\right) \\[5ex] y = \dfrac{3}{3} + \dfrac{2}{3} - \dfrac{1}{3} \\[5ex] y = \dfrac{3 + 2 - 1}{3} \\[5ex] y = \dfrac{5}{3}$
(4.) JAMB If the gradient of the curve $y = 2kx^2 + x + 1$ at $x = 1$ is $9$, find $k$

$A.\:\: 4 \\[3ex] B.\:\: 3 \\[3ex] C.\:\: 2 \\[3ex] D.\:\: 1 \\[3ex]$

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

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

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

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

$y = \dfrac{1}{x} = x^{-1} \\[5ex] \dfrac{dy}{dx} = -1 * x^{-1 - 1} \\[5ex] \dfrac{dy}{dx} = -x^{-2} \\[5ex] \dfrac{dy}{dx} = -\dfrac{1}{x^2}$
(7.) Find the derivative of $y = \sqrt{x}$ by first principle (by limits)

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 = $\sqrt{x + h} + \sqrt{x}$

$= \dfrac{\sqrt{x + h} - \sqrt{x}}{h} * \dfrac{\sqrt{x + h} + \sqrt{x}}{\sqrt{x + h} + \sqrt{x}} \\[5ex] = \dfrac{(\sqrt{x + h})^2 - (\sqrt{x})^2}{h(\sqrt{x + h} + \sqrt{x})} \\[5ex] = \dfrac{(x + h) - x}{h(\sqrt{x + h} + \sqrt{x})} \\[5ex] = \dfrac{x + h - x}{h(\sqrt{x + h} + \sqrt{x})} \\[5ex] = \dfrac{h}{h(\sqrt{x + h} + \sqrt{x})} \\[5ex] = \dfrac{1}{\sqrt{x + h} + \sqrt{x}} \\[5ex] f'(x) = \displaystyle{\lim_{h \to 0}}\: \dfrac{1}{\sqrt{x + h} + \sqrt{x}} \\[5ex] f'(x) = \dfrac{1}{\sqrt{x + 0} + \sqrt{x}} \\[5ex] f'(x) = \dfrac{1}{\sqrt{x} + \sqrt{x}} \\[5ex] f'(x) = \dfrac{1}{2\sqrt{x}} \\[5ex] \therefore \dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}}$
(8.) Find the derivative of $y = \sqrt{x}$ by second principle (by rules)

$y = \sqrt{x} \\[3ex] \sqrt{x} = x^{\dfrac{1}{2}} ...Law\:\:7...Exp \\[5ex] \rightarrow y = x^{\dfrac{1}{2}} \\[5ex] \dfrac{dy}{dx} = \dfrac{1}{2} * x^{\dfrac{1}{2} - 1} \\[5ex] \dfrac{dy}{dx} = \dfrac{1}{2} * x^{\dfrac{1}{2} - \dfrac{2}{2}} \\[5ex] \dfrac{dy}{dx} = \dfrac{1}{2} * x^-{\dfrac{1}{2}} \\[5ex] x^-{\dfrac{1}{2}} = \dfrac{1}{x^{\dfrac{1}{2}}} ...Law\:\:6...Exp \\[7ex] \dfrac{dy}{dx} = \dfrac{1}{2} * \dfrac{1}{x^{\dfrac{1}{2}}} \\[7ex] \dfrac{dy}{dx} = \dfrac{1}{2} * \dfrac{1}{\sqrt{x}} \\[7ex] \therefore \dfrac{dy}{dx} = \dfrac{1}{2\sqrt{x}}$
(9.) JAMB Differentiate $(2x + 5)^2(x - 4)$ with respect to $x$

$A.\:\: 4(2x + 5)(x - 4) \\[3ex] B.\:\: 4(2x + 5)(4x - 3) \\[3ex] C.\:\: (2x + 5)(2x - 13) \\[3ex] D.\:\: (2x + 5)(6x - 11) \\[3ex]$

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

$\dfrac{dy}{dx} = (2x + 5)(6x - 11)$
(10.) JAMB If $y = x\sin x$, find $\dfrac{dy}{dx}$ when $x = \dfrac{\pi}{2}$

$A.\:\: -\dfrac{\pi}{2} \\[5ex] B.\:\: -1 \\[3ex] C.\:\: 1 \\[3ex] D.\:\: \dfrac{\pi}{2} \\[5ex]$

Product Rule, Power Rule

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

$A.\:\: 2x + \dfrac{1}{x^2} \\[5ex] B.\:\: 2x + x^2 \\[3ex] C.\:\: 2x - \dfrac{1}{x^2} \\[5ex] D.\:\: 2x - x^2 \\[3ex]$

Power Rule

$y = x^2 - \dfrac{1}{x} \\[5ex] \dfrac{1}{x} = x^{-1}...Law\:\:6...Exp \\[3ex] y = x^2 - x^{-1} \\[3ex] \dfrac{dy}{dx} = 2 * x^{2 - 1} - -1 * x^{-1 - 1} \\[5ex] \dfrac{dy}{dx} = 2 * x^1 + 1 * x^{-2} \\[5ex] \dfrac{dy}{dx} = 2x + x^{-2} \\[5ex] \dfrac{dy}{dx} = 2x + \dfrac{1}{x^2}$
(12.) JAMB Find the slope of the curve $y = 2x^2 + 5x - 3$ at $(1, 4)$

$A.\:\: 9 \\[3ex] B.\:\: 7 \\[3ex] C.\:\: 6 \\[3ex] D.\:\: 4 \\[3ex]$

The slope is the derivative
Power Rule

$y = 2x^2 + 5x - 3 \\[3ex] y' = 4x + 5 \\[3ex] (1, 4) \rightarrow x = 1, y = 4 \\[3ex] y' = 4(1) + 5 \\[3ex] y' = 4 + 5 \\[3ex] y' = 9$
(13.) JAMB If $\dfrac{\pi}{2} \lt \theta \lt 2\pi$, find the maximum value of

$f(\theta) = \dfrac{4}{6 + 2\cos\theta}$

$A.\:\: \dfrac{1}{2} \\[5ex] B.\:\: 4 \\[3ex] C.\:\: 1 \\[3ex] D.\:\: \dfrac{2}{3} \\[5ex]$

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

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

$A.\:\: 2 \\[3ex] B.\:\: 4 \\[3ex] C.\:\: 6 \\[3ex] D.\:\: 0 \\[3ex]$
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.

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

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

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.) NSC Determine $\dfrac{dy}{dx}$ if:

$(17.1)\:\:y = 3x^3 + 6x^2 + x - 4 \\[3ex] (17.2)\:\: yx - y = 2x^2 - 2x;\:\:x \ne 1 \\[3ex]$

$(17.1) \\[3ex] y = 3x^3 + 6x^2 + x - 4 \\[3ex] \dfrac{dy}{dx} = 9x^2 + 12x + 1 \\[5ex] (17.2) \\[3ex] yx - y = 2x^2 - 2x \\[3ex] Factor\:\:by\:\:GCF \\[3ex] y(x - 1) = 2x(x - 1) \\[3ex] y = \dfrac{2x(x - 1)}{x - 1} \\[5ex] y = 2x \\[3ex] \dfrac{dy}{dx} = 2$
(18.) ATAR Use the quotient rule to show that

$\dfrac{d}{dx} \tan(x) = \dfrac{1}{\cos^2(x)}$

Quotient Rule

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

Chain Rule, Standard Derivatives

$y = \ln|\ln|\ln x|| \\[3ex] Let: \\[3ex] p = \ln x \:\:\:\: \dfrac{dp}{dx} = \dfrac{1}{x} \\[5ex] q = |p| \:\:\:\: \dfrac{dq}{dp} = \dfrac{|p|}{p} \\[5ex] r = \ln q \:\:\:\: \dfrac{dr}{dq} = \dfrac{1}{q} \\[5ex] s = |r| \:\:\:\: \dfrac{ds}{dr} = \dfrac{|r|}{r} \\[5ex] y = \ln s \:\:\:\: \dfrac{dy}{ds} = \dfrac{1}{s} \\[5ex] \dfrac{dy}{dx} = \dfrac{dp}{dx} * \dfrac{dq}{dp} * \dfrac{dr}{dq} * \dfrac{ds}{dr} * \dfrac{dy}{ds} \\[5ex] \dfrac{dy}{dx} = \dfrac{1}{x} * \dfrac{|p|}{p} * \dfrac{1}{q} * \dfrac{|r|}{r} * \dfrac{1}{s} \\[5ex] Substitute\:\:back \\[3ex] \dfrac{dy}{dx} = \dfrac{1}{x} * \dfrac{|\ln x|}{\ln x} * \dfrac{1}{|\ln x|} * \dfrac{|\ln q|}{\ln q} * \dfrac{1}{|\ln q|} \\[5ex] \dfrac{dy}{dx} = \dfrac{1}{x} * \dfrac{1}{\ln x} * \dfrac{1}{\ln q} \\[5ex] \dfrac{dy}{dx} = \dfrac{1}{x} * \dfrac{1}{\ln x} * \dfrac{1}{\ln |p|} \\[5ex] \dfrac{dy}{dx} = \dfrac{1}{x} * \dfrac{1}{\ln x} * \dfrac{1}{\ln |\ln x|} \\[5ex] \dfrac{dy}{dx} = \dfrac{1 * 1 * 1}{x * \ln x * \ln|\ln x|} \\[5ex] \dfrac{dy}{dx} = \dfrac{1}{x\ln x\ln|\ln x|}$
(20.) Find the derivative of $x^6y + y^6x = 9\:\:wrt\:\:x$

Implicit Differentiation, Product Rule, Power Rule

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

(21.) WASSCE Differentiate, with respect to $x$, $y = 3x^2 + 4x - 1$, from first principles.

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

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

Function of a Function Rule, Power Rule

$y = \left(3x - x^2\right)^5 \\[3ex] Let\;\; p = 3x - x^2 \;\;\; means\;\;that\;\; y = p^5 \\[3ex] \dfrac{dp}{dx} = 3 - 2x \\[5ex] \dfrac{dy}{dp} = 5p^4 \\[3ex] \dfrac{dy}{dx} = \dfrac{dy}{dp} * \dfrac{dp}{dx} ...Chain\;\;Rule \\[5ex] = 5p^4(3 - 2x) \\[3ex] = 5(3x - x^2)^4(3 - 2x) \\[3ex] = 5(3 - 2x)(3x - x^2)^4$
(23.) WASSCE Differentiate from first principles, with respect to $x$
$y = 3x^2 + 2x - 1$

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

$\underline{First\:\:Method:\:\:By\:\:Definition} \\[3ex] f(x) = 3x^2 + 2x - 1 \\[3ex] f(x + \Delta x) = 3(x + \Delta x)^2 + 2(x + \Delta x) - 1 \\[3ex] = 3[(x + \Delta x)(x + \Delta x)] + 2x + 2\Delta x - 1 \\[3ex] = 3[x^2 + x\Delta x + x\Delta x + (\Delta x)^2] + 2x + 2\Delta x - 1 \\[3ex] = 3[x^2 + 2x\Delta x + (\Delta x)^2] + 2x + 2\Delta x - 1 \\[3ex] = 3x^2 + 6x\Delta x + 3(\Delta x)^2 + 2x + 2\Delta x - 1 \\[3ex] = 3x^2 + 2x - 1 + 6x\Delta x + 2\Delta x + 3(\Delta x)^2 \\[5ex] f(x + \Delta x) - f(x) \\[3ex] = 3x^2 + 2x - 1 + 6x\Delta x + 2\Delta x + 3(\Delta x)^2 - (3x^2 + 2x - 1) \\[3ex] = 3x^2 + 2x - 1 + 6x\Delta x + 2\Delta x + 3(\Delta x)^2 - 3x^2 - 2x + 1 \\[3ex] = 6x\Delta x + 2\Delta x + 3(\Delta x)^2 \\[3ex] = \Delta x(6x + 2 + 3\Delta x) \\[5ex] \dfrac{f(x + \Delta x) - f(x)}{\Delta x} \\[5ex] = \dfrac{\Delta x(6x + 2 + 3\Delta x)}{\Delta x} \\[5ex] = 6x + 2 + 3\Delta x \\[5ex] f'(x) = \displaystyle{\lim_{\Delta x \to 0}}\:\: 6x + 2 + 3\Delta x \\[3ex] f'(x) = 6x + 2 + 3(0) \\[3ex] f'(x) = 6x + 2 \\[3ex] \underline{Second\:\:Method:\:\:By\:\:Formula} \\[3ex] f'(x) = \displaystyle{\lim_{h \to 0}}\: \dfrac{f(x + h) - f(x)}{h} \\[5ex] f(x) = 3x^2 + 2x - 1 \\[3ex] f(x + h) = 3(x + h)^2 + 2(x + h) - 1 \\[3ex] = 3[(x + h)(x + h)] + 2x + 2h - 1 \\[3ex] = 3(x^2 + hx + hx + h^2) + 2x + 2h - 1 \\[3ex] = 3(x^2 + 2hx + h^2) + 2x + 2h - 1 \\[3ex] = 3x^2 + 6hx + 3h^2 + 2x + 2h - 1 \\[3ex] \underline{Numerator} \\[3ex] f(x + h) - f(x) \\[3ex] = 3x^2 + 6hx + 3h^2 + 2x + 2h - 1 - (3x^2 + 2x - 1) \\[3ex] = 3x^2 + 6hx + 3h^2 + 2x + 2h - 1 - 3x^2 - 2x + 1 \\[3ex] = 6hx + 3h^2 + 2h \\[3ex] = h(6x + 3h + 2) \\[3ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{h(6x + 3h + 2)}{h} \\[5ex] = 6x + 3h + 2 \\[3ex] f'(x) = \displaystyle{\lim_{h \to 0}}\: 6x + 3h + 2 \\[3ex] f'(x) = 6x + 3(0) + 2 \\[3ex] f'(x) = 6x + 2$
(24.) NZQA Differentitate $y = \dfrac{\tan x}{x^3}$
$f(x) = \dfrac{\tan x}{x^3} \\[5ex] For\;\;f(x) = \dfrac{u}{v} \\[5ex] u = \tan x \\[3ex] u' = \sec^2 x \\[3ex] v = x^3 \\[3ex] v' = 3x^2 \\[3ex] f'(x) = \dfrac{vu' - uv'}{v^2} \\[5ex] \underline{Numerator} \\[3ex] vu' - uv' \\[3ex] = (x^3)(\sec^2 x) - (\tan x)(3x^2) \\[3ex] = x^3\sec^2 x - 3x^2\tan x \\[3ex] \underline{Denominator} \\[3ex] v^2 = (x^3)^2 = x^6 \\[3ex] \implies \\[3ex] f'(x) \\[3ex] = \dfrac{x^3\sec^2 x - 3x^2\tan x}{x^6} \\[5ex] = \dfrac{x^2(x\sec^2 x - 3\tan x)}{x^6} \\[5ex] = \dfrac{x\sec^2 x - 3\tan x}{x^4}$