Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN: 9781285741550
Author: James Stewart
Publisher: Cengage Learning
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Equations and inequalities describe the relationship between two mathematical expressions.
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To solve the problem of finding \(\frac{dy}{dt}\) given the function \(y = \ln \left( \frac{t^2 + 9}{\sqrt{8 - t}} \right)\), you will need to apply the chain rule and quotient rule from calculus. ### Solution Steps: 1. **Identify Parts of the Function:** - Let \(u = t^2 + 9\) and \(v = \sqrt{8 - t}\). - The function becomes \(y = \ln \left( \frac{u}{v} \right)\). 2. **Apply Logarithmic Differentiation:** - The derivative of \(\ln \left( \frac{u}{v} \right)\) is \(\frac{1}{\left( \frac{u}{v} \right)} \cdot \left( \frac{du/dt \cdot v - u \cdot dv/dt}{v^2} \right)\). 3. **Find Derivatives of \(u\) and \(v\):** - \(du/dt = 2t\). - \(dv/dt\) requires implicit differentiation: - \(v = (8-t)^{1/2}\), so \(dv/dt = -\frac{1}{2}(8-t)^{-1/2}\). 4. **Substitute into the Derivatives:** - Compute the derivative \(\frac{du}{v^2} = \frac{2t}{(8-t)}\). - Compute the second part: \(- \frac{u \cdot \frac{1}{2}(8-t)^{-3/2}}{v^2} = - \frac{(t^2 + 9)}{2(8-t)^{-3/2}}\). By solving these steps, you'll find the solution for \(\frac{dy}{dt}\). The crucial aspect is correctly applying derivative rules and simplifying the expressions. Once calculated, substitute the values back to get the explicit derivative \(\frac{dy}{dt} = \text{[your solution]}\).
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Transcribed Image Text:To solve the problem of finding \(\frac{dy}{dt}\) given the function \(y = \ln \left( \frac{t^2 + 9}{\sqrt{8 - t}} \right)\), you will need to apply the chain rule and quotient rule from calculus. ### Solution Steps: 1. **Identify Parts of the Function:** - Let \(u = t^2 + 9\) and \(v = \sqrt{8 - t}\). - The function becomes \(y = \ln \left( \frac{u}{v} \right)\). 2. **Apply Logarithmic Differentiation:** - The derivative of \(\ln \left( \frac{u}{v} \right)\) is \(\frac{1}{\left( \frac{u}{v} \right)} \cdot \left( \frac{du/dt \cdot v - u \cdot dv/dt}{v^2} \right)\). 3. **Find Derivatives of \(u\) and \(v\):** - \(du/dt = 2t\). - \(dv/dt\) requires implicit differentiation: - \(v = (8-t)^{1/2}\), so \(dv/dt = -\frac{1}{2}(8-t)^{-1/2}\). 4. **Substitute into the Derivatives:** - Compute the derivative \(\frac{du}{v^2} = \frac{2t}{(8-t)}\). - Compute the second part: \(- \frac{u \cdot \frac{1}{2}(8-t)^{-3/2}}{v^2} = - \frac{(t^2 + 9)}{2(8-t)^{-3/2}}\). By solving these steps, you'll find the solution for \(\frac{dy}{dt}\). The crucial aspect is correctly applying derivative rules and simplifying the expressions. Once calculated, substitute the values back to get the explicit derivative \(\frac{dy}{dt} = \text{[your solution]}\).
**Problem Statement:** Find \(\frac{dy}{dx}\) if \( y = \ln(x^3 \sqrt{x+1}) \). **Solution:** \[ \frac{dy}{dx} = \boxed{\phantom{x}} \] **Explanation:** To solve this problem, we need to differentiate the given function \( y = \ln(x^3 \sqrt{x+1}) \) with respect to \( x \). ### Steps for Differentiation: 1. **Use Logarithmic Properties:** Simplify the expression inside the logarithm using properties of logarithms: \[ y = \ln(x^3 \cdot (x+1)^{1/2}) = \ln(x^3) + \ln((x+1)^{1/2}) \] This can be further simplified to: \[ y = 3\ln(x) + \frac{1}{2}\ln(x+1) \] 2. **Differentiate:** Apply the derivative: - The derivative of \( 3\ln(x) \) is \(\frac{3}{x}\). - The derivative of \( \frac{1}{2}\ln(x+1) \) is \(\frac{1}{2} \cdot \frac{1}{x+1}\). Combine the derivatives: \[ \frac{dy}{dx} = \frac{3}{x} + \frac{1}{2(x+1)} \] This is the derivative of the function. Fill in the box with the expression obtained above.
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Transcribed Image Text:**Problem Statement:** Find \(\frac{dy}{dx}\) if \( y = \ln(x^3 \sqrt{x+1}) \). **Solution:** \[ \frac{dy}{dx} = \boxed{\phantom{x}} \] **Explanation:** To solve this problem, we need to differentiate the given function \( y = \ln(x^3 \sqrt{x+1}) \) with respect to \( x \). ### Steps for Differentiation: 1. **Use Logarithmic Properties:** Simplify the expression inside the logarithm using properties of logarithms: \[ y = \ln(x^3 \cdot (x+1)^{1/2}) = \ln(x^3) + \ln((x+1)^{1/2}) \] This can be further simplified to: \[ y = 3\ln(x) + \frac{1}{2}\ln(x+1) \] 2. **Differentiate:** Apply the derivative: - The derivative of \( 3\ln(x) \) is \(\frac{3}{x}\). - The derivative of \( \frac{1}{2}\ln(x+1) \) is \(\frac{1}{2} \cdot \frac{1}{x+1}\). Combine the derivatives: \[ \frac{dy}{dx} = \frac{3}{x} + \frac{1}{2(x+1)} \] This is the derivative of the function. Fill in the box with the expression obtained above.
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