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Original
2026-01-01
Modified
2026-02-28
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<p>We can derive the derivative of x⁴ using proofs.</p>
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<p>We can derive the derivative of x⁴ using proofs.</p>
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<p>To show this, we will use algebraic manipulation along with the rules of differentiation.</p>
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<p>To show this, we will use algebraic manipulation 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</p>
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<p>By First Principle</p>
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<p>Using Power Rule</p>
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<p>Using Power Rule</p>
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<p>Using Product Rule</p>
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<p>Using Product Rule</p>
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<p>We will now demonstrate that the differentiation of x⁴ results in 4x³ using the above-mentioned methods:</p>
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<p>We will now demonstrate that the differentiation of x⁴ results in 4x³ using the above-mentioned methods:</p>
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<p>By First Principle</p>
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<p>By First Principle</p>
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<p>The derivative of x⁴ 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>The derivative of x⁴ 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 x⁴ using the first principle, we will consider f(x) = x⁴.</p>
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<p>To find the derivative of x⁴ using the first principle, we will consider f(x) = x⁴.</p>
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<p>Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)</p>
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<p>Its derivative can be expressed as the following limit. f'(x) = limₕ→₀ [f(x + h) - f(x)] / h … (1)</p>
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<p>Given that f(x) = x⁴, we write f(x + h) = (x + h)⁴.</p>
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<p>Given that f(x) = x⁴, we write f(x + h) = (x + h)⁴.</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [(x + h)⁴ - x⁴] / h = limₕ→₀ [x⁴ + 4x³h + 6x²h² + 4xh³ + h⁴ - x⁴] / h = limₕ→₀ [4x³h + 6x²h² + 4xh³ + h⁴] / h = limₕ→₀ [4x³ + 6x²h + 4xh² + h³]</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [(x + h)⁴ - x⁴] / h = limₕ→₀ [x⁴ + 4x³h + 6x²h² + 4xh³ + h⁴ - x⁴] / h = limₕ→₀ [4x³h + 6x²h² + 4xh³ + h⁴] / h = limₕ→₀ [4x³ + 6x²h + 4xh² + h³]</p>
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<p>As h approaches 0, we have, f'(x) = 4x³.</p>
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<p>As h approaches 0, we have, f'(x) = 4x³.</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<p>Using Power Rule</p>
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<p>Using Power Rule</p>
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<p>To prove the differentiation of x⁴ using the power rule,</p>
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<p>To prove the differentiation of x⁴ using the power rule,</p>
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<p>We use the formula: d/dx (xⁿ) = n*xⁿ⁻¹</p>
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<p>We use the formula: d/dx (xⁿ) = n*xⁿ⁻¹</p>
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<p>For x⁴, n = 4, so: d/dx (x⁴) = 4*x³</p>
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<p>For x⁴, n = 4, so: d/dx (x⁴) = 4*x³</p>
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<p>This is a straightforward application of the power rule.</p>
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<p>This is a straightforward application of the power rule.</p>
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<p>Using Product Rule</p>
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<p>Using Product Rule</p>
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<p>We will now prove the derivative of x⁴ using the<a>product</a>rule.</p>
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<p>We will now prove the derivative of x⁴ using the<a>product</a>rule.</p>
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<p>The step-by-step process is demonstrated below:</p>
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<p>The step-by-step process is demonstrated below:</p>
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<p>Here, we rewrite x⁴ as x*x*x*x.</p>
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<p>Here, we rewrite x⁴ as x*x*x*x.</p>
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<p>Using the product rule formula:</p>
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<p>Using the product rule formula:</p>
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<p>d/dx [u.v] = u'.v + u.v'</p>
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<p>d/dx [u.v] = u'.v + u.v'</p>
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<p>Consider u = x and v = x³. u' = d/dx (x) = 1 v' = d/dx (x³) = 3x²</p>
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<p>Consider u = x and v = x³. u' = d/dx (x) = 1 v' = d/dx (x³) = 3x²</p>
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<p>Using the product rule: d/dx (x⁴) = u'.v + u.v' = 1*x³ + x*3x² = x³ + 3x³ = 4x³</p>
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<p>Using the product rule: d/dx (x⁴) = u'.v + u.v' = 1*x³ + x*3x² = x³ + 3x³ = 4x³</p>
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<p>Thus: d/dx (x⁴) = 4x³.</p>
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<p>Thus: d/dx (x⁴) = 4x³.</p>
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