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2026-01-01
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<p>129 Learners</p>
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<p>Last updated on<strong>October 8, 2025</strong></p>
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<p>Last updated on<strong>October 8, 2025</strong></p>
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<p>We explore the derivatives of transcendental functions, such as \( \tan(x) \). The derivative, \( \sec^2(x) \), serves as a tool to measure how the tangent function changes with respect to small changes in \( x \). Derivatives are crucial in calculating changes, like profit or loss, in real-life situations. We will delve into the derivatives of transcendental functions.</p>
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<p>We explore the derivatives of transcendental functions, such as \( \tan(x) \). The derivative, \( \sec^2(x) \), serves as a tool to measure how the tangent function changes with respect to small changes in \( x \). Derivatives are crucial in calculating changes, like profit or loss, in real-life situations. We will delve into the derivatives of transcendental functions.</p>
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<h2>What is the Derivative of Transcendental Functions?</h2>
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<h2>What is the Derivative of Transcendental Functions?</h2>
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<p>Understanding the derivative<a>of</a>transcendental<a>functions</a>, like ( tan(x) ), is essential in<a>calculus</a>. The derivative of ( tan(x) ) is commonly represented as ( frac{d}{dx} (tan x) ) or ( (tan x)' ), and its value is ( sec^2(x) ). This indicates that the tangent function is differentiable within its domain.</p>
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<p>Understanding the derivative<a>of</a>transcendental<a>functions</a>, like ( tan(x) ), is essential in<a>calculus</a>. The derivative of ( tan(x) ) is commonly represented as ( frac{d}{dx} (tan x) ) or ( (tan x)' ), and its value is ( sec^2(x) ). This indicates that the tangent function is differentiable within its domain.</p>
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<p>Key concepts include: </p>
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<p>Key concepts include: </p>
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<p>Tangent Function: ( tan(x) = frac{sin(x)}{cos(x)} ). </p>
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<p>Tangent Function: ( tan(x) = frac{sin(x)}{cos(x)} ). </p>
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<p>Quotient Rule: A rule for differentiating ( tan(x) ) due to its form ( frac{sin(x)}{cos(x)} ). </p>
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<p>Quotient Rule: A rule for differentiating ( tan(x) ) due to its form ( frac{sin(x)}{cos(x)} ). </p>
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<p>Secant Function: ( sec(x) = frac{1}{cos(x)} ).</p>
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<p>Secant Function: ( sec(x) = frac{1}{cos(x)} ).</p>
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<h2>Derivative of Transcendental Functions Formula</h2>
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<h2>Derivative of Transcendental Functions Formula</h2>
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<p>The derivative of ( tan(x) ) can be denoted as ( frac{d}{dx} (tan x) ) or ( (tan x)' ).</p>
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<p>The derivative of ( tan(x) ) can be denoted as ( frac{d}{dx} (tan x) ) or ( (tan x)' ).</p>
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<p>The<a>formula</a>for differentiating ( tan(x) ) is: [ frac{d}{dx} (tan x) = sec2 x ]</p>
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<p>The<a>formula</a>for differentiating ( tan(x) ) is: [ frac{d}{dx} (tan x) = sec2 x ]</p>
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<p>This formula is valid for all ( x ) where ( cos(x) neq 0 ).</p>
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<p>This formula is valid for all ( x ) where ( cos(x) neq 0 ).</p>
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<h2>Proofs of the Derivative of Transcendental Functions</h2>
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<h2>Proofs of the Derivative of Transcendental Functions</h2>
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<p>The derivative of ( tan(x) ) can be derived using various proofs involving trigonometric identities and differentiation rules.</p>
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<p>The derivative of ( tan(x) ) can be derived using various proofs involving trigonometric identities and differentiation rules.</p>
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<p>Methods include: </p>
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<p>Methods include: </p>
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<p>By First Principle: Using the limit of the difference<a>quotient</a>.</p>
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<p>By First Principle: Using the limit of the difference<a>quotient</a>.</p>
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<p>Using Chain Rule: Applying chain differentiation techniques. </p>
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<p>Using Chain Rule: Applying chain differentiation techniques. </p>
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<p>Using Product Rule: Breaking down products within functions.</p>
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<p>Using Product Rule: Breaking down products within functions.</p>
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<h2>By First Principle</h2>
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<h2>By First Principle</h2>
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<p>To prove using the First Principle, express the derivative as the limit of the difference quotient for ( f(x) = tan x ): [ f'(x) = lim_{h to 0} frac{tan(x + h) - tan x}{h} ] [ = lim_{h to 0} frac{frac{sin(x + h)}{cos(x + h)} - frac{ sin x}{\cos x}}{h} \] \[ = \lim_{h \to 0} \frac{\sin(x + h)\cos x - \cos(x + h)\sin x}{h \cos x \cos(x + h)} \] Using the identity \( \sin A \cos B - \cos A \sin B = \sin(A - B) \), it becomes: \[ = \lim_{h \to 0} \frac{\sin h}{h \cos x \cos(x + h)} \] Using \( \lim_{h \to 0} \frac{\sin h}{h} = 1 \), it simplifies to \( \frac{1}{\cos^2 x} \). Thus: \[ f'(x) = \sec^2 x \]</p>
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<p>To prove using the First Principle, express the derivative as the limit of the difference quotient for ( f(x) = tan x ): [ f'(x) = lim_{h to 0} frac{tan(x + h) - tan x}{h} ] [ = lim_{h to 0} frac{frac{sin(x + h)}{cos(x + h)} - frac{ sin x}{\cos x}}{h} \] \[ = \lim_{h \to 0} \frac{\sin(x + h)\cos x - \cos(x + h)\sin x}{h \cos x \cos(x + h)} \] Using the identity \( \sin A \cos B - \cos A \sin B = \sin(A - B) \), it becomes: \[ = \lim_{h \to 0} \frac{\sin h}{h \cos x \cos(x + h)} \] Using \( \lim_{h \to 0} \frac{\sin h}{h} = 1 \), it simplifies to \( \frac{1}{\cos^2 x} \). Thus: \[ f'(x) = \sec^2 x \]</p>
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<h2>Using Chain Rule</h2>
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<h2>Using Chain Rule</h2>
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<p>For the chain rule: \[ \tan x = \frac{\sin x}{\cos x} \] Using \( f(x) = \sin x \) and \( g(x) = \cos x \), apply the quotient rule: \[ \frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} \] Substituting values gives: \[ \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x \]</p>
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<p>For the chain rule: \[ \tan x = \frac{\sin x}{\cos x} \] Using \( f(x) = \sin x \) and \( g(x) = \cos x \), apply the quotient rule: \[ \frac{d}{dx} \left(\frac{f(x)}{g(x)}\right) = \frac{f'(x)g(x) - f(x)g'(x)}{g(x)^2} \] Substituting values gives: \[ \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x \]</p>
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<h2>Using Product Rule</h2>
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<h2>Using Product Rule</h2>
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<p>Express \( \tan x \) as \( \sin x \cdot (\cos x)^{-1} \) and apply the<a>product</a>rule. Given \( u = \sin x \) and \( v = (\cos x)^{-1} \), use: \[ \frac{d}{dx}[u \cdot v] = u' \cdot v + u \cdot v' \] With \( u' = \cos x \) and \( v' = \frac{\sin x}{\cos^2 x} \): \[ \frac{d}{dx} (\tan x) = \cos x \cdot (\cos x)^{-1} + \sin x \cdot \frac{\sin x}{\cos^2 x} \] Simplifying: \[ = \sec^2 x \]</p>
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<p>Express \( \tan x \) as \( \sin x \cdot (\cos x)^{-1} \) and apply the<a>product</a>rule. Given \( u = \sin x \) and \( v = (\cos x)^{-1} \), use: \[ \frac{d}{dx}[u \cdot v] = u' \cdot v + u \cdot v' \] With \( u' = \cos x \) and \( v' = \frac{\sin x}{\cos^2 x} \): \[ \frac{d}{dx} (\tan x) = \cos x \cdot (\cos x)^{-1} + \sin x \cdot \frac{\sin x}{\cos^2 x} \] Simplifying: \[ = \sec^2 x \]</p>
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<h2>Higher-Order Derivatives of Transcendental Functions</h2>
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<h2>Higher-Order Derivatives of Transcendental Functions</h2>
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<p>Higher-order derivatives involve differentiating a function<a>multiple</a>times. They can be complex but provide deeper insights. For a function \( f(x) \), the first derivative \( f'(x) \) indicates the<a>rate</a>of change.</p>
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<p>Higher-order derivatives involve differentiating a function<a>multiple</a>times. They can be complex but provide deeper insights. For a function \( f(x) \), the first derivative \( f'(x) \) indicates the<a>rate</a>of change.</p>
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<p>The second derivative \( f''(x) \) shows the rate of change of the rate of change, and this pattern continues. For \( \tan(x) \), successive derivatives highlight how changes evolve over time.</p>
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<p>The second derivative \( f''(x) \) shows the rate of change of the rate of change, and this pattern continues. For \( \tan(x) \), successive derivatives highlight how changes evolve over time.</p>
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<h2>Special Cases</h2>
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<h2>Special Cases</h2>
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<p>At \( x = \frac{\pi}{2} \), the derivative is undefined due to the vertical asymptote of \( \tan(x) \).</p>
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<p>At \( x = \frac{\pi}{2} \), the derivative is undefined due to the vertical asymptote of \( \tan(x) \).</p>
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<p>At \( x = 0 \), the derivative of \( \tan x \) is \( \sec^2(0) = 1 \).</p>
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<p>At \( x = 0 \), the derivative of \( \tan x \) is \( \sec^2(0) = 1 \).</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of Transcendental Functions</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of Transcendental Functions</h2>
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<p>Errors often occur when differentiating \( \tan(x) \). Understanding correct processes can prevent common mistakes:</p>
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<p>Errors often occur when differentiating \( \tan(x) \). Understanding correct processes can prevent common mistakes:</p>
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<h3>Problem 1</h3>
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<h3>Problem 1</h3>
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<p>Calculate the derivative of \( \tan x \cdot \sec^2 x \).</p>
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<p>Calculate the derivative of \( \tan x \cdot \sec^2 x \).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Given \( f(x) = \tan x \cdot \sec^2 x \), apply the product rule: \[ f'(x) = u'v + uv' \] Here, \( u = \tan x \) and \( v = \sec^2 x \). Differentiate each: \[ u' = \sec^2 x, \quad v' = 2 \sec^2 x \tan x \] Substitute: \[ f'(x) = (\sec^2 x) \cdot (\sec^2 x) + (\tan x) \cdot (2 \sec^2 x \tan x) \] Simplify: \[ f'(x) = \sec^4 x + 2 \sec^2 x \tan^2 x \]</p>
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<p>Given \( f(x) = \tan x \cdot \sec^2 x \), apply the product rule: \[ f'(x) = u'v + uv' \] Here, \( u = \tan x \) and \( v = \sec^2 x \). Differentiate each: \[ u' = \sec^2 x, \quad v' = 2 \sec^2 x \tan x \] Substitute: \[ f'(x) = (\sec^2 x) \cdot (\sec^2 x) + (\tan x) \cdot (2 \sec^2 x \tan x) \] Simplify: \[ f'(x) = \sec^4 x + 2 \sec^2 x \tan^2 x \]</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>The derivative is calculated using the product rule, dividing the function into parts, differentiating them, and combining the results.</p>
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<p>The derivative is calculated using the product rule, dividing the function into parts, differentiating them, and combining the results.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 2</h3>
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<h3>Problem 2</h3>
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<p>A roller coaster's height is modeled by \( y = \tan(x) \). If \( x = \frac{\pi}{6} \) radians, find the slope.</p>
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<p>A roller coaster's height is modeled by \( y = \tan(x) \). If \( x = \frac{\pi}{6} \) radians, find the slope.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Given \( y = \tan(x) \), differentiate: \[ \frac{dy}{dx} = \sec^2(x) \] Substitute \( x = \frac{\pi}{6} \): \[ \sec^2\left(\frac{\pi}{6}\right) = 1 + \tan^2\left(\frac{\pi}{6}\right) \] Since \( \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \): \[ \sec^2\left(\frac{\pi}{6}\right) = \frac{4}{3} \] The slope is \( \frac{4}{3} \).</p>
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<p>Given \( y = \tan(x) \), differentiate: \[ \frac{dy}{dx} = \sec^2(x) \] Substitute \( x = \frac{\pi}{6} \): \[ \sec^2\left(\frac{\pi}{6}\right) = 1 + \tan^2\left(\frac{\pi}{6}\right) \] Since \( \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \): \[ \sec^2\left(\frac{\pi}{6}\right) = \frac{4}{3} \] The slope is \( \frac{4}{3} \).</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>The slope at \( x = \frac{\pi}{6} \) is calculated by substituting into the derivative \( \sec^2(x) \).</p>
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<p>The slope at \( x = \frac{\pi}{6} \) is calculated by substituting into the derivative \( \sec^2(x) \).</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 3</h3>
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<h3>Problem 3</h3>
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<p>Derive the second derivative of \( y = \tan(x) \).</p>
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<p>Derive the second derivative of \( y = \tan(x) \).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>First derivative: \[ \frac{dy}{dx} = \sec^2(x) \] Differentiate again for the second derivative: \[ \frac{d^2y}{dx^2} = \frac{d}{dx}[\sec^2(x)] \] Using the chain rule: \[ = 2 \sec(x) \cdot \sec(x) \tan(x) \] Simplify: \[ = 2 \sec^2(x) \tan(x) \]</p>
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<p>First derivative: \[ \frac{dy}{dx} = \sec^2(x) \] Differentiate again for the second derivative: \[ \frac{d^2y}{dx^2} = \frac{d}{dx}[\sec^2(x)] \] Using the chain rule: \[ = 2 \sec(x) \cdot \sec(x) \tan(x) \] Simplify: \[ = 2 \sec^2(x) \tan(x) \]</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>The second derivative involves differentiating the first derivative using the product and chain rules.</p>
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<p>The second derivative involves differentiating the first derivative using the product and chain rules.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 4</h3>
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<h3>Problem 4</h3>
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<p>Prove: \( \frac{d}{dx} (\tan^2(x)) = 2 \tan(x) \sec^2(x) \).</p>
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<p>Prove: \( \frac{d}{dx} (\tan^2(x)) = 2 \tan(x) \sec^2(x) \).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Using the chain rule, consider \( y = \tan^2(x) = (\tan(x))^2 \): \[ \frac{dy}{dx} = 2 \tan(x) \cdot \frac{d}{dx}[\tan(x)] \] Since \( \frac{d}{dx}[\tan(x)] = \sec^2(x) \): \[ \frac{dy}{dx} = 2 \tan(x) \cdot \sec^2(x) \] Hence proved.</p>
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<p>Using the chain rule, consider \( y = \tan^2(x) = (\tan(x))^2 \): \[ \frac{dy}{dx} = 2 \tan(x) \cdot \frac{d}{dx}[\tan(x)] \] Since \( \frac{d}{dx}[\tan(x)] = \sec^2(x) \): \[ \frac{dy}{dx} = 2 \tan(x) \cdot \sec^2(x) \] Hence proved.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>The chain rule is applied to differentiate \( y = \tan^2(x) \), substituting the derivative of \( \tan(x) \).</p>
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<p>The chain rule is applied to differentiate \( y = \tan^2(x) \), substituting the derivative of \( \tan(x) \).</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h3>Problem 5</h3>
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<h3>Problem 5</h3>
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<p>Solve: \( \frac{d}{dx} \left(\frac{\tan x}{x}\right) \).</p>
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<p>Solve: \( \frac{d}{dx} \left(\frac{\tan x}{x}\right) \).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Using the quotient rule: \[ \frac{d}{dx} \left(\frac{\tan x}{x}\right) = \frac{x \cdot \frac{d}{dx}(\tan x) - \tan x \cdot \frac{d}{dx}(x)}{x^2} \] Substitute \( \frac{d}{dx}(\tan x) = \sec^2 x \): \[ = \frac{x \sec^2 x - \tan x}{x^2} \]</p>
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<p>Using the quotient rule: \[ \frac{d}{dx} \left(\frac{\tan x}{x}\right) = \frac{x \cdot \frac{d}{dx}(\tan x) - \tan x \cdot \frac{d}{dx}(x)}{x^2} \] Substitute \( \frac{d}{dx}(\tan x) = \sec^2 x \): \[ = \frac{x \sec^2 x - \tan x}{x^2} \]</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>The function is differentiated using the quotient rule, simplifying the expression to get the final derivative.</p>
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<p>The function is differentiated using the quotient rule, simplifying the expression to get the final derivative.</p>
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<p>Well explained 👍</p>
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<p>Well explained 👍</p>
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<h2>FAQs on the Derivative of Transcendental Functions</h2>
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<h2>FAQs on the Derivative of Transcendental Functions</h2>
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<h3>1.Find the derivative of \( \tan x \).</h3>
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<h3>1.Find the derivative of \( \tan x \).</h3>
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<p>Using the quotient rule for \( \tan x = \frac{\sin x}{\cos x} \), the derivative is: \[ \frac{d}{dx} (\tan x) = \sec^2 x \]</p>
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<p>Using the quotient rule for \( \tan x = \frac{\sin x}{\cos x} \), the derivative is: \[ \frac{d}{dx} (\tan x) = \sec^2 x \]</p>
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<h3>2.Can we use the derivative of \( \tan x \) in real life?</h3>
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<h3>2.Can we use the derivative of \( \tan x \) in real life?</h3>
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<p>Yes, derivatives of transcendental functions are used to calculate rates of change in various fields like physics, engineering, and economics.</p>
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<p>Yes, derivatives of transcendental functions are used to calculate rates of change in various fields like physics, engineering, and economics.</p>
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<h3>3.Is it possible to take the derivative of \( \tan x \) at \( x = \frac{\pi}{2} \)?</h3>
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<h3>3.Is it possible to take the derivative of \( \tan x \) at \( x = \frac{\pi}{2} \)?</h3>
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<p>No, at \( x = \frac{\pi}{2} \), \( \tan x \) is undefined, making the derivative impossible at these points.</p>
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<p>No, at \( x = \frac{\pi}{2} \), \( \tan x \) is undefined, making the derivative impossible at these points.</p>
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<h3>4.What rule is used to differentiate \( \frac{\tan x}{x} \)?</h3>
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<h3>4.What rule is used to differentiate \( \frac{\tan x}{x} \)?</h3>
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<p>The quotient rule is used to differentiate \( \frac{\tan x}{x} \), resulting in: \[ \frac{x \sec^2 x - \tan x}{x^2} \]</p>
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<p>The quotient rule is used to differentiate \( \frac{\tan x}{x} \), resulting in: \[ \frac{x \sec^2 x - \tan x}{x^2} \]</p>
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<h3>5.Are the derivatives of \( \tan x \) and \( \tan^{-1} x \) the same?</h3>
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<h3>5.Are the derivatives of \( \tan x \) and \( \tan^{-1} x \) the same?</h3>
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<p>No, they differ. The derivative of \( \tan x \) is \( \sec^2 x \), while the derivative of \( \tan^{-1} x \) is \( \frac{1}{1 + x^2} \).</p>
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<p>No, they differ. The derivative of \( \tan x \) is \( \sec^2 x \), while the derivative of \( \tan^{-1} x \) is \( \frac{1}{1 + x^2} \).</p>
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<h2>Important Glossaries for the Derivative of Transcendental Functions</h2>
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<h2>Important Glossaries for the Derivative of Transcendental Functions</h2>
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<ul><li><strong>Derivative:</strong>Measures how a function changes as its input changes.</li>
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<ul><li><strong>Derivative:</strong>Measures how a function changes as its input changes.</li>
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</ul><ul><li><strong>Tangent Function:</strong>The ratio of the opposite to the adjacent side in a right triangle, denoted as \( \tan x \). </li>
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</ul><ul><li><strong>Tangent Function:</strong>The ratio of the opposite to the adjacent side in a right triangle, denoted as \( \tan x \). </li>
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</ul><ul><li><strong>Secant Function:</strong>Reciprocal of the cosine function, represented as \( \sec x \).</li>
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</ul><ul><li><strong>Secant Function:</strong>Reciprocal of the cosine function, represented as \( \sec x \).</li>
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</ul><ul><li><strong>First Principle:</strong>Method of finding derivatives using the limit of the difference quotient.</li>
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</ul><ul><li><strong>First Principle:</strong>Method of finding derivatives using the limit of the difference quotient.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule for differentiating compositions of functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A rule for differentiating compositions of functions.</li>
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</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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</ul><p>What Is Calculus? 🔢 | Easy Tricks, Limits & 🎯 Fun Learning for Kids | ✨BrightCHAMPS Math</p>
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<p>▶</p>
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<p>▶</p>
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<h2>Jaskaran Singh Saluja</h2>
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<h2>Jaskaran Singh Saluja</h2>
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<h3>About the Author</h3>
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<h3>About the Author</h3>
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<p>Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.</p>
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<p>Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.</p>
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<h3>Fun Fact</h3>
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<h3>Fun Fact</h3>
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<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>
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<p>: He loves to play the quiz with kids through algebra to make kids love it.</p>