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2026-01-01
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<p>Last updated on<strong>September 15, 2025</strong></p>
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<p>Last updated on<strong>September 15, 2025</strong></p>
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<p>We use the derivative of ln(3t), which is 1/t, as a measuring tool for how the natural logarithm function changes in response to a slight change in t. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of ln(3t) in detail.</p>
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<p>We use the derivative of ln(3t), which is 1/t, as a measuring tool for how the natural logarithm function changes in response to a slight change in t. Derivatives help us calculate profit or loss in real-life situations. We will now talk about the derivative of ln(3t) in detail.</p>
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<h2>What is the Derivative of ln(3t)?</h2>
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<h2>What is the Derivative of ln(3t)?</h2>
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<p>We now understand the derivative of ln(3t). It is commonly represented as d/dt (ln(3t)) or (ln(3t))', and its value is 1/t. The<a>function</a>ln(3t) has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:</p>
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<p>We now understand the derivative of ln(3t). It is commonly represented as d/dt (ln(3t)) or (ln(3t))', and its value is 1/t. The<a>function</a>ln(3t) has a clearly defined derivative, indicating it is differentiable within its domain. The key concepts are mentioned below:</p>
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<p>Natural Logarithm Function: (ln(x)) is the inverse of the exponential function e^x.</p>
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<p>Natural Logarithm Function: (ln(x)) is the inverse of the exponential function e^x.</p>
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<p>Chain Rule: A rule for differentiating ln(3t) because it consists of a composition<a>of functions</a>.</p>
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<p>Chain Rule: A rule for differentiating ln(3t) because it consists of a composition<a>of functions</a>.</p>
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<p>Reciprocal Function: The derivative involves 1/t.</p>
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<p>Reciprocal Function: The derivative involves 1/t.</p>
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<h2>Derivative of ln(3t) Formula</h2>
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<h2>Derivative of ln(3t) Formula</h2>
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<p>The derivative of ln(3t) can be denoted as d/dt (ln(3t)) or (ln(3t))'.</p>
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<p>The derivative of ln(3t) can be denoted as d/dt (ln(3t)) or (ln(3t))'.</p>
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<p>The<a>formula</a>we use to differentiate ln(3t) is: d/dt (ln(3t)) = 1/t (or) (ln(3t))' = 1/t</p>
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<p>The<a>formula</a>we use to differentiate ln(3t) is: d/dt (ln(3t)) = 1/t (or) (ln(3t))' = 1/t</p>
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<p>The formula applies to all t where t > 0.</p>
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<p>The formula applies to all t where t > 0.</p>
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<h2>Proofs of the Derivative of ln(3t)</h2>
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<h2>Proofs of the Derivative of ln(3t)</h2>
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<p>We can derive the derivative of ln(3t) using proofs. To show this, we will use logarithmic identities along with the rules of differentiation. There are several methods we use to prove this, such as:</p>
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<p>We can derive the derivative of ln(3t) using proofs. To show this, we will use logarithmic identities along with the rules of differentiation. 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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<li>Using the properties of<a>logarithms</a></li>
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<li>Using the properties of<a>logarithms</a></li>
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</ol><p>We will now demonstrate that the differentiation of ln(3t) results in 1/t using the above-mentioned methods:</p>
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</ol><p>We will now demonstrate that the differentiation of ln(3t) results in 1/t 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 ln(3t) 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 ln(3t) 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 ln(3t) using the first principle, we will consider f(t) = ln(3t). Its derivative can be expressed as the following limit. f'(t) = limₕ→₀ [f(t + h) - f(t)] / h … (1)</p>
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<p>To find the derivative of ln(3t) using the first principle, we will consider f(t) = ln(3t). Its derivative can be expressed as the following limit. f'(t) = limₕ→₀ [f(t + h) - f(t)] / h … (1)</p>
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<p>Given that f(t) = ln(3t), we write f(t + h) = ln(3(t + h)).</p>
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<p>Given that f(t) = ln(3t), we write f(t + h) = ln(3(t + h)).</p>
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<p>Substituting these into<a>equation</a>(1), f'(t) = limₕ→₀ [ln(3(t + h)) - ln(3t)] / h = limₕ→₀ ln[(3(t + h))/(3t)] / h = limₕ→₀ ln[1 + h/t] / h</p>
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<p>Substituting these into<a>equation</a>(1), f'(t) = limₕ→₀ [ln(3(t + h)) - ln(3t)] / h = limₕ→₀ ln[(3(t + h))/(3t)] / h = limₕ→₀ ln[1 + h/t] / h</p>
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<p>For small h, ln(1 + h/t) ≈ h/t, f'(t) = limₕ→₀ (h/t) / h = 1/t.</p>
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<p>For small h, ln(1 + h/t) ≈ h/t, f'(t) = limₕ→₀ (h/t) / h = 1/t.</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 ln(3t) using the chain rule, Consider f(t) = ln(u) and u(t) = 3t.</p>
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<p>To prove the differentiation of ln(3t) using the chain rule, Consider f(t) = ln(u) and u(t) = 3t.</p>
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<p>By the chain rule: d/dt [ln(u)] = (1/u) * (du/dt) Substitute u = 3t, d/dt (ln(3t)) = (1/3t) * 3 = 1/t.</p>
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<p>By the chain rule: d/dt [ln(u)] = (1/u) * (du/dt) Substitute u = 3t, d/dt (ln(3t)) = (1/3t) * 3 = 1/t.</p>
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<h3>Using Properties of Logarithms</h3>
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<h3>Using Properties of Logarithms</h3>
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<p>We use the property ln(ab) = ln(a) + ln(b). ln(3t) = ln(3) + ln(t)</p>
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<p>We use the property ln(ab) = ln(a) + ln(b). ln(3t) = ln(3) + ln(t)</p>
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<p>The derivative of ln(3) is 0 since it is a<a>constant</a>. d/dt (ln(t)) = 1/t.</p>
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<p>The derivative of ln(3) is 0 since it is a<a>constant</a>. d/dt (ln(t)) = 1/t.</p>
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<p>So, d/dt (ln(3t)) = 1/t.</p>
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<p>So, d/dt (ln(3t)) = 1/t.</p>
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<h2>Higher-Order Derivatives of ln(3t)</h2>
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<h2>Higher-Order Derivatives of ln(3t)</h2>
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<p>When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky. To understand them better, think of a car where the speed changes (first derivative) and the<a>rate</a>at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like ln(3t).</p>
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<p>When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives. Higher-order derivatives can be a little tricky. To understand them better, think of a car where the speed changes (first derivative) and the<a>rate</a>at which the speed changes (second derivative) also changes. Higher-order derivatives make it easier to understand functions like ln(3t).</p>
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<p>For the first derivative of a function, we write f′(t), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(t). Similarly, the third derivative, f′′′(t) is the result of the second derivative and this pattern continues.</p>
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<p>For the first derivative of a function, we write f′(t), which indicates how the function changes or its slope at a certain point. The second derivative is derived from the first derivative, which is denoted using f′′(t). Similarly, the third derivative, f′′′(t) is the result of the second derivative and this pattern continues.</p>
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<p>For the nth Derivative of ln(3t), we generally use f^(n)(t) for the nth derivative of a function f(t), which tells us the change in the rate of change (continuing for higher-order derivatives).</p>
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<p>For the nth Derivative of ln(3t), we generally use f^(n)(t) for the nth derivative of a function f(t), which tells us the change in the rate of change (continuing for higher-order derivatives).</p>
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<h2>Special Cases:</h2>
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<h2>Special Cases:</h2>
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<p>When t = 0, the derivative is undefined because ln(3t) has a vertical asymptote there. When t = 1, the derivative of ln(3t) = 1/1, which is 1/3.</p>
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<p>When t = 0, the derivative is undefined because ln(3t) has a vertical asymptote there. When t = 1, the derivative of ln(3t) = 1/1, which is 1/3.</p>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of ln(3t)</h2>
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<h2>Common Mistakes and How to Avoid Them in Derivatives of ln(3t)</h2>
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<p>Students frequently make mistakes when differentiating ln(3t). These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:</p>
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<p>Students frequently make mistakes when differentiating ln(3t). These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:</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 ln(3t) · e^t.</p>
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<p>Calculate the derivative of ln(3t) · e^t.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Here, we have f(t) = ln(3t) · e^t. Using the product rule, f'(t) = u′v + uv′. In the given equation, u = ln(3t) and v = e^t.</p>
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<p>Here, we have f(t) = ln(3t) · e^t. Using the product rule, f'(t) = u′v + uv′. In the given equation, u = ln(3t) and v = e^t.</p>
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<p>Let’s differentiate each term, u′ = d/dt (ln(3t)) = 1/t. v′ = d/dt (e^t) = e^t.</p>
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<p>Let’s differentiate each term, u′ = d/dt (ln(3t)) = 1/t. v′ = d/dt (e^t) = e^t.</p>
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<p>Substituting into the given equation, f'(t) = (1/t) · e^t + ln(3t) · e^t.</p>
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<p>Substituting into the given equation, f'(t) = (1/t) · e^t + ln(3t) · e^t.</p>
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<p>Let’s simplify terms to get the final answer, f'(t) = e^t/t + ln(3t) · e^t.</p>
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<p>Let’s simplify terms to get the final answer, f'(t) = e^t/t + ln(3t) · e^t.</p>
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<p>Thus, the derivative of the specified function is e^t/t + ln(3t) · e^t.</p>
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<p>Thus, the derivative of the specified function is e^t/t + ln(3t) · e^t.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find the derivative of the given function by dividing the function into two parts. The first step is finding its derivative and then combining them using the product rule to get the final result.</p>
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<p>We find the derivative of the given function by dividing the function into two parts. The first step is finding its derivative and then combining them using the product rule to get the final result.</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 company's revenue is represented by the function y = ln(3t), where y represents the revenue at time t. If t = 2 years, measure the rate of change of revenue with respect to time.</p>
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<p>A company's revenue is represented by the function y = ln(3t), where y represents the revenue at time t. If t = 2 years, measure the rate of change of revenue with respect to time.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>We have y = ln(3t) (revenue function)...(1)</p>
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<p>We have y = ln(3t) (revenue function)...(1)</p>
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<p>Now, we will differentiate the equation (1)</p>
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<p>Now, we will differentiate the equation (1)</p>
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<p>Take the derivative ln(3t): dy/dt = 1/t. Given t = 2 (substitute this into the derivative)</p>
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<p>Take the derivative ln(3t): dy/dt = 1/t. Given t = 2 (substitute this into the derivative)</p>
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<p>dy/dt = 1/2.</p>
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<p>dy/dt = 1/2.</p>
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<p>Hence, we get the rate of change of revenue at t = 2 years as 1/2.</p>
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<p>Hence, we get the rate of change of revenue at t = 2 years as 1/2.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We find the rate of change of revenue at t = 2 years as 1/2, which means that at this point, the revenue increases at half the rate of time.</p>
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<p>We find the rate of change of revenue at t = 2 years as 1/2, which means that at this point, the revenue increases at half the rate of time.</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 the function y = ln(3t).</p>
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<p>Derive the second derivative of the function y = ln(3t).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>The first step is to find the first derivative, dy/dt = 1/t...(1)</p>
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<p>The first step is to find the first derivative, dy/dt = 1/t...(1)</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dt² = d/dt (1/t).</p>
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<p>Now we will differentiate equation (1) to get the second derivative: d²y/dt² = d/dt (1/t).</p>
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<p>Here we use the derivative of reciprocal function, d²y/dt² = -1/t².</p>
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<p>Here we use the derivative of reciprocal function, d²y/dt² = -1/t².</p>
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<p>Therefore, the second derivative of the function y = ln(3t) is -1/t².</p>
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<p>Therefore, the second derivative of the function y = ln(3t) is -1/t².</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>We use the step-by-step process, where we start with the first derivative. Using the derivative of the reciprocal function, we differentiate 1/t. We then simplify the terms to find the final answer.</p>
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<p>We use the step-by-step process, where we start with the first derivative. Using the derivative of the reciprocal function, we differentiate 1/t. We then simplify the terms to find the final answer.</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: d/dt (ln(3t)²) = 2 ln(3t)/t.</p>
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<p>Prove: d/dt (ln(3t)²) = 2 ln(3t)/t.</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>Let’s start using the chain rule: Consider y = (ln(3t))².</p>
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<p>Let’s start using the chain rule: Consider y = (ln(3t))².</p>
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<p>To differentiate, we use the chain rule: dy/dt = 2 ln(3t) · d/dt [ln(3t)].</p>
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<p>To differentiate, we use the chain rule: dy/dt = 2 ln(3t) · d/dt [ln(3t)].</p>
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<p>Since the derivative of ln(3t) is 1/t, dy/dt = 2 ln(3t) · 1/t.</p>
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<p>Since the derivative of ln(3t) is 1/t, dy/dt = 2 ln(3t) · 1/t.</p>
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<p>Substituting y = ln(3t)², d/dt (ln(3t)²) = 2 ln(3t)/t.</p>
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<p>Substituting y = ln(3t)², d/dt (ln(3t)²) = 2 ln(3t)/t.</p>
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<p>Hence proved.</p>
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<p>Hence proved.</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace ln(3t) with its derivative. As a final step, we substitute y = ln(3t)² to derive the equation.</p>
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<p>In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace ln(3t) with its derivative. As a final step, we substitute y = ln(3t)² to derive the equation.</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: d/dt (ln(3t)/t).</p>
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<p>Solve: d/dt (ln(3t)/t).</p>
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<p>Okay, lets begin</p>
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<p>Okay, lets begin</p>
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<p>To differentiate the function, we use the quotient rule: d/dt (ln(3t)/t) = (d/dt (ln(3t)) · t - ln(3t) · d/dt(t)) / t².</p>
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<p>To differentiate the function, we use the quotient rule: d/dt (ln(3t)/t) = (d/dt (ln(3t)) · t - ln(3t) · d/dt(t)) / t².</p>
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<p>We will substitute d/dt (ln(3t)) = 1/t and d/dt(t) = 1. (1/t · t - ln(3t) · 1) / t². = (1 - ln(3t)) / t².</p>
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<p>We will substitute d/dt (ln(3t)) = 1/t and d/dt(t) = 1. (1/t · t - ln(3t) · 1) / t². = (1 - ln(3t)) / t².</p>
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<p>Therefore, d/dt (ln(3t)/t) = (1 - ln(3t)) / t².</p>
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<p>Therefore, d/dt (ln(3t)/t) = (1 - ln(3t)) / t².</p>
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<h3>Explanation</h3>
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<h3>Explanation</h3>
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<p>In this process, we differentiate the given function using the quotient rule. As a final step, we simplify the equation to obtain the final result.</p>
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<p>In this process, we differentiate the given function using the quotient rule. As a final step, we simplify the equation to obtain the final result.</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 ln(3t)</h2>
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<h2>FAQs on the Derivative of ln(3t)</h2>
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<h3>1.Find the derivative of ln(3t).</h3>
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<h3>1.Find the derivative of ln(3t).</h3>
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<p>Using the chain rule to differentiate ln(3t), we get: d/dt (ln(3t)) = 1/t (simplified).</p>
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<p>Using the chain rule to differentiate ln(3t), we get: d/dt (ln(3t)) = 1/t (simplified).</p>
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<h3>2.Can we use the derivative of ln(3t) in real life?</h3>
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<h3>2.Can we use the derivative of ln(3t) in real life?</h3>
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<p>Yes, we can use the derivative of ln(3t) in real life to calculate the rate of change of any process, especially in fields such as mathematics, finance, and physics.</p>
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<p>Yes, we can use the derivative of ln(3t) in real life to calculate the rate of change of any process, especially in fields such as mathematics, finance, and physics.</p>
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<h3>3.Is it possible to take the derivative of ln(3t) at the point where t = 0?</h3>
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<h3>3.Is it possible to take the derivative of ln(3t) at the point where t = 0?</h3>
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<p>No, t = 0 is a point where ln(3t) is undefined, so it is impossible to take the derivative at this point (since the function does not exist there).</p>
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<p>No, t = 0 is a point where ln(3t) is undefined, so it is impossible to take the derivative at this point (since the function does not exist there).</p>
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<h3>4.What rule is used to differentiate ln(3t)/t?</h3>
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<h3>4.What rule is used to differentiate ln(3t)/t?</h3>
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<p>We use the quotient rule to differentiate ln(3t)/t: d/dt (ln(3t)/t) = (1 - ln(3t)) / t².</p>
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<p>We use the quotient rule to differentiate ln(3t)/t: d/dt (ln(3t)/t) = (1 - ln(3t)) / t².</p>
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<h3>5.Are the derivatives of ln(3t) and ln(t) the same?</h3>
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<h3>5.Are the derivatives of ln(3t) and ln(t) the same?</h3>
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<p>No, they are different. The derivative of ln(3t) is 1/t, while the derivative of ln(t) is also 1/t, but the context and application involving the constant<a>multiplier</a>differ in interpretation.</p>
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<p>No, they are different. The derivative of ln(3t) is 1/t, while the derivative of ln(t) is also 1/t, but the context and application involving the constant<a>multiplier</a>differ in interpretation.</p>
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<h2>Important Glossaries for the Derivative of ln(3t)</h2>
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<h2>Important Glossaries for the Derivative of ln(3t)</h2>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the given function changes in response to a slight change in its variable.</li>
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<ul><li><strong>Derivative:</strong>The derivative of a function indicates how the given function changes in response to a slight change in its variable.</li>
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</ul><ul><li><strong>Natural Logarithm:</strong>The natural logarithm is the logarithm to the base e, where e is the constant approximately equal to 2.71828.</li>
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</ul><ul><li><strong>Natural Logarithm:</strong>The natural logarithm is the logarithm to the base e, where e is the constant approximately equal to 2.71828.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A fundamental rule in calculus used to differentiate compositions of functions.</li>
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</ul><ul><li><strong>Chain Rule:</strong>A fundamental rule in calculus used to differentiate compositions of functions.</li>
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</ul><ul><li><strong>Reciprocal Function:</strong>A function that is the reciprocal of another function, often resulting in 1/x.</li>
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</ul><ul><li><strong>Reciprocal Function:</strong>A function that is the reciprocal of another function, often resulting in 1/x.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule for differentiating functions that are ratios of two other functions.</li>
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</ul><ul><li><strong>Quotient Rule:</strong>A rule for differentiating functions that are ratios of two other 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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<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>