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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 sqrt(3x) using proofs. To show this, we will use the rules of differentiation and chain rule.</p>
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<p>We can derive the derivative of sqrt(3x) using proofs. To show this, we will use the rules of differentiation and chain rule.</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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<ol><li>By First Principle</li>
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<ol><li>By First Principle</li>
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<li>Using Chain Rule</li>
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<li>Using Chain Rule</li>
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</ol><p>We will now demonstrate that the differentiation of sqrt(3x) results in 3/(2sqrt(3x)) using the above-mentioned methods:</p>
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</ol><p>We will now demonstrate that the differentiation of sqrt(3x) results in 3/(2sqrt(3x)) using the above-mentioned methods:</p>
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<h3>By First Principle</h3>
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<h3>By First Principle</h3>
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<p>The derivative of sqrt(3x) 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 sqrt(3x) 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 sqrt(3x) using the first principle, we will consider f(x) = sqrt(3x). 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>To find the derivative of sqrt(3x) using the first principle, we will consider f(x) = sqrt(3x). 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) = sqrt(3x), we write f(x + h) = sqrt(3(x + h)).</p>
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<p>Given that f(x) = sqrt(3x), we write f(x + h) = sqrt(3(x + h)).</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [sqrt(3(x + h)) - sqrt(3x)] / h</p>
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<p>Substituting these into<a>equation</a>(1), f'(x) = limₕ→₀ [sqrt(3(x + h)) - sqrt(3x)] / h</p>
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<p>Multiply and divide by the<a>conjugate</a>: = limₕ→₀ [(sqrt(3(x + h)) - sqrt(3x)) * (sqrt(3(x + h)) + sqrt(3x))] / [h * (sqrt(3(x + h)) + sqrt(3x))] = limₕ→₀ [3(x + h) - 3x] / [h * (sqrt(3(x + h)) + sqrt(3x))] = limₕ→₀ [3h] / [h * (sqrt(3(x + h)) + sqrt(3x))]</p>
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<p>Multiply and divide by the<a>conjugate</a>: = limₕ→₀ [(sqrt(3(x + h)) - sqrt(3x)) * (sqrt(3(x + h)) + sqrt(3x))] / [h * (sqrt(3(x + h)) + sqrt(3x))] = limₕ→₀ [3(x + h) - 3x] / [h * (sqrt(3(x + h)) + sqrt(3x))] = limₕ→₀ [3h] / [h * (sqrt(3(x + h)) + sqrt(3x))]</p>
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<p>Cancel h: = limₕ→₀ 3 / [sqrt(3(x + h)) + sqrt(3x)] = 3 / [2sqrt(3x)]</p>
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<p>Cancel h: = limₕ→₀ 3 / [sqrt(3(x + h)) + sqrt(3x)] = 3 / [2sqrt(3x)]</p>
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<p>Hence, proved.</p>
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<p>Hence, proved.</p>
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<h3>Using Chain Rule</h3>
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<h3>Using Chain Rule</h3>
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<p>To prove the differentiation of sqrt(3x) using the chain rule, We use the formula: Sqrt(3x) = (3x)^(1/2) Let u = 3x</p>
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<p>To prove the differentiation of sqrt(3x) using the chain rule, We use the formula: Sqrt(3x) = (3x)^(1/2) Let u = 3x</p>
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<p>Then, sqrt(3x) = u^(1/2)</p>
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<p>Then, sqrt(3x) = u^(1/2)</p>
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<p>By the chain rule: d/dx [u^(n)] = n * u^(n-1) * (du/dx)… (1)</p>
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<p>By the chain rule: d/dx [u^(n)] = n * u^(n-1) * (du/dx)… (1)</p>
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<p>Let’s substitute u = 3x, n = 1/2 into equation (1), d/dx (sqrt(3x)) = (1/2) * (3x)^(-1/2) * d/dx (3x) = (1/2) * (3x)^(-1/2) * 3 = 3/(2sqrt(3x))</p>
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<p>Let’s substitute u = 3x, n = 1/2 into equation (1), d/dx (sqrt(3x)) = (1/2) * (3x)^(-1/2) * d/dx (3x) = (1/2) * (3x)^(-1/2) * 3 = 3/(2sqrt(3x))</p>
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<p>Thus: d/dx (sqrt(3x)) = 3/(2sqrt(3x)).</p>
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<p>Thus: d/dx (sqrt(3x)) = 3/(2sqrt(3x)).</p>
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