Derivative (First Principle)

Section 3.1 Derivative (First Principle)

In calculus, the derivative of a function \(f(x)\) measures how the function’s output changes as its input changes. In other words, the instantaneous rate of change or the slope of the tangent line to the curve at a point. The derivative is denoted as: \(f'(x)\text{.}\) The first principle (also called the limit definition or difference quotient) is the foundational way to find the derivative by using the formula below.

\begin{equation*} f'(x)=\lim_{h\to 0} \frac{f(x+h)-f(x)}{h} \end{equation*}

However, there is shortcut rules to find the derivative of a function \(f(x).\) If \(y=f(x)\) then,

\begin{equation*} f'(x) = \frac{\,dy}{\,dx} \end{equation*}

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Subsection 3.1.1 Derivative using First Principle

Example 3.1.1.

Find the derivative of \(\sin x\) using first principle.

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Solution.

Let \(f(x) = \sin x \) then,

\begin{align*} f'(x) \amp = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ \amp = \lim_{h \to 0} \frac{\sin (x+h)-\sin (x)}{h}\\ \amp = \lim_{h \to 0} \frac{2\cos\frac{(x+h+x)}{2}\cdot \sin\frac{(x+h-x)}{2} }{h} \end{align*}

[Trigonometric Identities]

\begin{align*} \amp = \lim_{h \to 0} \frac{2\cos\frac{(2x+h)}{2}\cdot \sin\frac{(h)}{2} }{\frac{h}{2}\times 2}\\ \amp = \lim_{h \to 0} \frac{2\cos\left(\frac{2x+h}{2}\right)}{2} \cdot \lim_{h \to 0}\frac{\sin\left(\frac{h}{2}\right) }{\left(\frac{h}{2}\right)}\\ \amp = \cos \frac{2x}{2}\times 1 \\ \therefore \, f'(x) \amp = \cos x \end{align*}

[Some Standard Limits]

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Example 3.1.2.

Find the derivative of \(\tan\,x\) using first principle.

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Solution.

Let \(f(x) = \tan\,x \) then,

\begin{align*} f'(x) \amp = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\\ \amp = \lim_{h \to 0} \frac{\tan (x+h)-\tan (x)}{h}\\ \amp = \lim_{h \to 0} \frac{1}{h} \left[ \frac{\sin (x+h)}{\cos (x+h)}-\frac{\sin x}{\cos x}\right] \end{align*}

[Trigonometric Identities]

\begin{align*} \amp = \lim_{h \to 0} \frac{1}{h} \left[ \frac{\sin (x+h)\cos x-\sin x \cos\,(x+h)}{\cos (x+h)\cos x}\right]\\ \amp = \lim_{h \to 0} \frac{1}{h} \left[ \frac{\sin (x+h-x)}{\cos (x+h)\cos x}\right]\\ \amp = \lim_{h \to 0} \frac{\sin h}{h} \cdot \lim_{h \to 0} \frac{1}{\cos (x+h)\cos x} = 1\times \frac{1}{\cos^2 x}\\ \therefore \, f'(x) \amp = \sec^2 x. \end{align*}

[Some Standard Limits]

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Example 3.1.3.

Find the derivative of \(\sin^2 x\) using first principle.

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Solution.

Let \(f(x) = \sin^2 x \) then,

\begin{align*} f'(x) \amp = \lim_{h \to 0} \frac{\sin^2 (x+h)-\sin^2 (x)}{h}\\ \amp = \lim_{h \to 0} \frac{1}{h} \left[ \frac{1-\cos\,2(x+h)}{2}-\frac{1-\cos \,2x}{2}\right]\\ \amp = \lim_{h \to 0} \frac{1}{2h} \left[ 1-\cos\,2(x+h)-1+\cos\,2x\right]\\ \amp = \lim_{h \to 0} \frac{1}{2h} \left[ \cos\,2x-\cos\,2(x+h)\right] \end{align*}

[Trigonometric Identities]

\begin{align*} \amp = \lim_{h \to 0} \frac{1}{2h} \left[-2 \sin\{\frac{2x+2x+2h}{2}\} \cdot \sin\{\frac{2x-2x-2h}{2}\}\right]\\ \amp = \lim_{h \to 0} \frac{1}{\cancel{2}h} \left[-\cancel{2} \sin\{\frac{4x+2h}{2}\} \cdot \sin\{\frac{\cancel{2}h}{\cancel{2}}\}\right] \end{align*}

[Some Standard Limits]

\begin{align*} \amp = \lim_{h \to 0} \sin\{\frac{4x+2h}{2}\} \cdot \lim_{h \to 0} \sin\{\frac{h}{h}\} \end{align*}

\begin{equation*} \therefore \, f'(x) = \sin \,2x. \end{equation*}

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Example 3.1.4.

Find the derivative of \(x^n\) using first principle.

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Solution.

Let \(f(x) = x^n \) then,

\begin{align*} f'(x) \amp = \lim_{h \to 0} \frac{(x+h)^n-x^n}{h}\\ \amp = \lim_{h \to 0} \frac{1}{h} \left[ {}^{n}C_{0} x^n + {}^{n}C_{1} x^{n-1}h +{}^{n}C_{2} x^{n-2}h^2 +\cdots +{}^{n}C_{n} x^{n-n}h^n -x^n \right]\\ \amp = \lim_{h \to 0} \frac{1}{h} \left[ \cancel{x^n} + n x^{n-1}h +\frac{n(n-1)}{2} x^{n-2}h^2 +\cdots + x^{0}h^n -\cancel{x^n} \right]\\ \amp = \lim_{h \to 0} \left[ n x^{n-1} +\frac{n(n-1)}{2} x^{n-2}h +\cdots + h^{n-1} \right]\\ \therefore \, f'(x) \amp = n x^{n-1}. \end{align*}

[Binomial Theorem and Combination and Permutation]

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Example 3.1.5.

Find the derivative of \(\log\,x\) using first principle.

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Solution.

Let \(f(x) = \log\,x \) then,

\begin{align*} f'(x) \amp = \lim_{h \to 0} \frac{\log\,(x+h)-\log\,x}{h}\\ \amp = \lim_{h \to 0} \frac{1}{h} \log\,\{ \frac{x+h}{x}\}\\ \amp = \lim_{h \to 0} \frac{\log\,\{ 1+\frac{h}{x}\}}{\frac{h}{x}x} \\ \amp = \frac{1}{x} \lim_{h \to 0} \frac{\log\,\{ 1+\frac{h}{x}\}}{\frac{h}{x}} \\ \therefore \, f'(x) \amp = \frac{1}{x}. \end{align*}

[Logarithimic and Exponential]

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Example 3.1.6.

Find the derivative of \(e^x\) using first principle.

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Solution.

Let \(f(x) = e^x\) then,

\begin{align*} f'(x) \amp = \lim_{h \to 0} \frac{e^{(x+h)}-e^x}{h}\\ \amp = \lim_{h \to 0} \frac{e^x\cdot e^h-e^x}{h}\\ \amp = \lim_{h \to 0} \frac{e^x[e^h-1]}{h}\\ \therefore \, f'(x) \amp = e^x. \end{align*}

[Logarithimic and Exponential]

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