Solved Examples: Derivatives by Limits

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

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

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

(4.) Determine the derivative of $x - 2x^3$ by limits

We can solve this question using at least two approaches.
Use any approach you prefer.

$ \underline{\text{First Approach: By Formula: Limit of the Difference Quotient}} \\[3ex] f'(x) = \displaystyle{\lim_{h \to 0}}\: \dfrac{f(x + h) - f(x)}{h} \\[5ex] f(x) = x - 2x^3 \\[3ex] f(x + h) = (x + h) - 2(x + h)^3 \\[3ex] ....................................................... \\[3ex] \text{Let us the Pascal's Triangle to expand } (x + h)^3 \\[3ex] \begin{array}{c} 1 \\[2ex] 1 \quad 1 \\[2ex] 1 \quad 2 \quad 1 \\[2ex] 1 \quad 3 \quad 3 \quad 1 \\[2ex] \end{array} \\[10ex] \implies \\[3ex] (x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3 \\[3ex] ....................................................... \\[3ex] = x + h - 2[x^3 + 3x^2h + 3xh^2 + h^3] \\[3ex] = x + h - 2x^3 - 6x^2h - 6xh^2 - 2h^3 \\[3ex] \underline{Numerator} \\[3ex] f(x + h) - f(x) \\[3ex] = x + h - 2x^3 - 6x^2h - 6xh^2 - 2h^3 - (x - 2x^3) \\[3ex] = x + h - 2x^3 - 6x^2h - 6xh^2 - 2h^3 - x + 2x^3 \\[3ex] = h - 6x^2h - 6xh^2 - 2h^3 \\[3ex] = h(1 - 6x^2 - 6xh - 2h^2) \\[3ex] \dfrac{Numerator}{Denominator} \\[5ex] = \dfrac{h(1 - 6x^2 - 6xh - 2h^2)}{h} \\[5ex] = 1 - 6x^2 - 6xh - 2h^2 \\[3ex] f'(x) = \displaystyle{\lim_{h \to 0}}\: [1 - 6x^2 - 6xh - 2h^2] \\[3ex] f'(x) = 1 - 6x^2 - 6x(0) - 2(0)^2 \\[3ex] f'(x) = 1 - 6x^2 \\[5ex] \underline{\text{Second Approach: By Definition}} \\[3ex] 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 $

(5.) Differentiate $y = \dfrac{1}{x}$ wrt $x$ by limits.

(6.) Find the derivative of $y = \sqrt{x}$ by first principle (by limits)

We can solve this question using at least two approaches.
Use any approach you prefer.

$ \underline{\text{First Approach: By Formula: Limit of the Difference Quotient}} \\[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}}\: \left[\dfrac{1}{\sqrt{x + h} + \sqrt{x}}\right] \\[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}} \\[5ex] \underline{\text{Second Approach: 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} \\[5ex] = \displaystyle{\lim_{\Delta x \to 0}} \left[\dfrac{1}{\sqrt{x + \Delta x} + \sqrt{x}}\right] \\[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}} $

(7.) SACE Consider the function $f(x) = \dfrac{x}{x - 2}$, where x ≠ 2.

(a.) Use first principles to find $f'(x)$
(b.) Hence, or otherwise, state the slope of the tangent to $f(x)$ when x = 6.

$ (a.) \\[3ex] f(x) = \dfrac{x}{x - 2}\;\;\; x \ne 6 \\[5ex] \text{Difference Quotient, DQ} = \dfrac{f(x + h) - f(x)}{h} \\[5ex] f(x + h) = \dfrac{x + h}{(x + h) - 2} \\[5ex] f(x + h) = \dfrac{x + h}{x + h - 2} \\[5ex] \underline{\text{Numerator of DQ}} \\[3ex] f(x + h) - f(x) \\[3ex] = \dfrac{x + h}{x + h - 2} -\dfrac{x}{x - 2} \\[5ex] = \dfrac{(x + h)(x - 2) - x(x + h - 2)}{(x + h - 2)(x - 2)} \\[5ex] = \dfrac{x^2 - 2x + xh - 2h - x^2 - xh + 2x}{(x + h - 2)(x - 2)} \\[5ex] = \dfrac{-2h}{(x + h - 2)(x - 2)} \\[5ex] \underline{\text{Numerator and Denominator of DQ}} \\[3ex] DQ = [f(x + h) - f(x)] \div h \\[3ex] DQ = \dfrac{-2h}{(x + h - 2)(x - 2)} * \dfrac{1}{h} \\[5ex] DQ = \dfrac{-2}{(x + h - 2)(x - 2)} \\[5ex] \underline{Derivative} \\[3ex] \text{Derivative},\; f'(x) = \displaystyle{\lim_{h \to 0}} DQ \\[3ex] f'(x) = \displaystyle{\lim_{h \to 0}} \dfrac{-2}{(x + h - 2)(x - 2)} \\[5ex] f'(x) = \dfrac{-2}{(x + 0 - 2)(x - 2)} \\[5ex] f'(x) = \dfrac{-2}{(x - 2)(x - 2)} \\[5ex] f'(x) = \dfrac{-2}{(x - 2)^2} \\[5ex] (b.) \\[3ex] \text{slope of the tangent to f(x) when x = 6 is the derivative of f(x) at x = 6} \\[3ex] = \left.\dfrac{dy}{dx}\right|_{x = 6} \\[5ex] = \dfrac{-2}{(6 - 2)^2} \\[5ex] = \dfrac{-2}{4^2} \\[5ex] = \dfrac{-2}{16} \\[5ex] = -\dfrac{1}{8} $

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