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2026-01-01
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2026-02-28
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<p>We can derive the derivative of 1/cos(x) using proofs.</p>
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<p>We can derive the derivative of 1/cos(x) using proofs.</p>
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<p>To show this, we will use the trigonometric identities along with the rules of differentiation.</p>
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<p>To show this, we will use the trigonometric identities along with the rules of differentiation.</p>
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<p>There are several methods we use to prove this, such as:</p>
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<p>There are several methods we use to prove this, such as:</p>
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<p>By First Principle Using Chain Rule Using Product Rule We will now demonstrate that the differentiation of 1/cos(x) results in sec(x)tan(x) using the above-mentioned methods: By First Principle The derivative of 1/cos(x) can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>.</p>
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<p>By First Principle Using Chain Rule Using Product Rule We will now demonstrate that the differentiation of 1/cos(x) results in sec(x)tan(x) using the above-mentioned methods: By First Principle The derivative of 1/cos(x) can be proved using the First Principle, which expresses the derivative as the limit of the difference<a>quotient</a>.</p>
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<p>To find the derivative of 1/cos(x) using the first principle, we will consider f(x) = 1/cos(x).</p>
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<p>To find the derivative of 1/cos(x) using the first principle, we will consider f(x) = 1/cos(x).</p>
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<p>Its derivative can be expressed as the following limit.</p>
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<p>Its derivative can be expressed as the following limit.</p>
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<p>f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)</p>
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<p>f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)</p>
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<p>Given that f(x) = 1/cos(x), we write f(x + h) = 1/cos(x + h).</p>
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<p>Given that f(x) = 1/cos(x), we write f(x + h) = 1/cos(x + h).</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [1/cos(x + h) - 1/cos(x)] / h = limₕ→₀ [(cos(x) - cos(x + h)) / (cos(x)cos(x + h))] / h = limₕ→₀ [(-sin(x + h/2)sin(h/2)) / (cos(x)cos(x + h))]/ h</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [1/cos(x + h) - 1/cos(x)] / h = limₕ→₀ [(cos(x) - cos(x + h)) / (cos(x)cos(x + h))] / h = limₕ→₀ [(-sin(x + h/2)sin(h/2)) / (cos(x)cos(x + h))]/ h</p>
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<p>Using the identity for sin A - sin B, f'(x) = limₕ→₀ [-sin(x + h/2)sin(h/2)] / [h cos(x)cos(x + h)] = limₕ→₀ [-sin(x + h/2)(h/2)/(h/2)cos(x)cos(x + h)]</p>
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<p>Using the identity for sin A - sin B, f'(x) = limₕ→₀ [-sin(x + h/2)sin(h/2)] / [h cos(x)cos(x + h)] = limₕ→₀ [-sin(x + h/2)(h/2)/(h/2)cos(x)cos(x + h)]</p>
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<p>Using limit formula, limₕ→₀ (sin(h/2))/(h/2) = 1.</p>
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<p>Using limit formula, limₕ→₀ (sin(h/2))/(h/2) = 1.</p>
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<p>f'(x) = -sin(x) / cos²(x) As the reciprocal of cosine is secant, and using tan(x) = sin(x)/cos(x), we have, f'(x) = sec(x)tan(x). Hence, proved.</p>
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<p>f'(x) = -sin(x) / cos²(x) As the reciprocal of cosine is secant, and using tan(x) = sin(x)/cos(x), we have, f'(x) = sec(x)tan(x). Hence, proved.</p>
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<p>Using Chain Rule To prove the differentiation of 1/cos(x) using the chain rule, Consider y = 1/cos(x).</p>
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<p>Using Chain Rule To prove the differentiation of 1/cos(x) using the chain rule, Consider y = 1/cos(x).</p>
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<p>Let u = cos(x), then y = 1/u.</p>
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<p>Let u = cos(x), then y = 1/u.</p>
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<p>Using the chain rule: dy/dx = dy/du * du/dx.</p>
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<p>Using the chain rule: dy/dx = dy/du * du/dx.</p>
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<p>We have dy/du = -1/u² and du/dx = -sin(x).</p>
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<p>We have dy/du = -1/u² and du/dx = -sin(x).</p>
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<p>Therefore, dy/dx = -1/cos²(x) * -sin(x) = sin(x)/cos²(x).</p>
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<p>Therefore, dy/dx = -1/cos²(x) * -sin(x) = sin(x)/cos²(x).</p>
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<p>Using the identity tan(x) = sin(x)/cos(x), we have, dy/dx = sec(x)tan(x). Using Product Rule We will now prove the derivative of 1/cos(x) using the<a>product</a>rule.</p>
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<p>Using the identity tan(x) = sin(x)/cos(x), we have, dy/dx = sec(x)tan(x). Using Product Rule We will now prove the derivative of 1/cos(x) using the<a>product</a>rule.</p>
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<p>Let y = sec(x) = 1/cos(x) = (cos(x))⁻¹.</p>
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<p>Let y = sec(x) = 1/cos(x) = (cos(x))⁻¹.</p>
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<p>Using the product rule formula: d/dx [u.v] = u'.v + u.v'. Let u = 1 and v = (cos(x))⁻¹.</p>
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<p>Using the product rule formula: d/dx [u.v] = u'.v + u.v'. Let u = 1 and v = (cos(x))⁻¹.</p>
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<p>Then v' = -sin(x)/(cos(x))². Using the product rule, dy/dx = 0 + 1 * (-sin(x)/(cos(x))²) = -sin(x)/cos²(x) = sec(x)tan(x).</p>
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<p>Then v' = -sin(x)/(cos(x))². Using the product rule, dy/dx = 0 + 1 * (-sin(x)/(cos(x))²) = -sin(x)/cos²(x) = sec(x)tan(x).</p>
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