Pregunta
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Use the Chain Rule to find \( \frac{d z}{d t} \), where \( z=3 x \sin y, x=t^{4} \), and \( y=3 t^{2} \). \( \frac{d z}{d t}=\square \) (Type an expression using tas the variable. Use parentheses to clearly denote the argument of each function.)

Ask by Peters Cole. in the United States
Mar 17,2025

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\( \frac{d z}{d t} = 12t^{3}\sin(3t^{2}) + 18t^{5}\cos(3t^{2}) \)

Solución

To find \( \frac{d z}{d t} \) using the Chain Rule, we need to differentiate the function \( z = 3x \sin y \) with respect to \( t \), where \( x = t^4 \) and \( y = 3t^2 \). Given: - \( z = 3x \sin y \) - \( x = t^4 \) - \( y = 3t^2 \) Using the Chain Rule, we can find \( \frac{d z}{d t} \) by differentiating \( z \) with respect to \( t \) and then applying the chain rule to account for the intermediate variables \( x \) and \( y \). Let's start by finding \( \frac{d z}{d t} \) using the Chain Rule. Evaluate the derivative by following steps: - step0: Solution: \(\frac{d}{dt}\left(3t^{4}\sin\left(3t^{2}\right)\right)\) - step1: Use differentiation rules: \(\frac{d}{dt}\left(3\right)\times t^{4}\sin\left(3t^{2}\right)+3\times \frac{d}{dt}\left(t^{4}\right)\times \sin\left(3t^{2}\right)+3t^{4}\times \frac{d}{dt}\left(\sin\left(3t^{2}\right)\right)\) - step2: Find the derivative: \(0\times t^{4}\sin\left(3t^{2}\right)+3\times \frac{d}{dt}\left(t^{4}\right)\times \sin\left(3t^{2}\right)+3t^{4}\times \frac{d}{dt}\left(\sin\left(3t^{2}\right)\right)\) - step3: Calculate: \(0+3\times \frac{d}{dt}\left(t^{4}\right)\times \sin\left(3t^{2}\right)+3t^{4}\times \frac{d}{dt}\left(\sin\left(3t^{2}\right)\right)\) - step4: Calculate: \(0+3\times 4t^{3}\sin\left(3t^{2}\right)+3t^{4}\times \frac{d}{dt}\left(\sin\left(3t^{2}\right)\right)\) - step5: Calculate: \(0+12t^{3}\sin\left(3t^{2}\right)+3t^{4}\times \frac{d}{dt}\left(\sin\left(3t^{2}\right)\right)\) - step6: Calculate: \(0+12t^{3}\sin\left(3t^{2}\right)+3t^{4}\times 6t\cos\left(3t^{2}\right)\) - step7: Calculate: \(0+12t^{3}\sin\left(3t^{2}\right)+18t^{5}\cos\left(3t^{2}\right)\) - step8: Remove 0: \(12t^{3}\sin\left(3t^{2}\right)+18t^{5}\cos\left(3t^{2}\right)\) The derivative of \( z \) with respect to \( t \) using the Chain Rule is \( \frac{d z}{d t} = 12t^{3}\sin(3t^{2}) + 18t^{5}\cos(3t^{2}) \).

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The Deep Dive

To find \( \frac{dz}{dt} \) using the Chain Rule, we can start by computing the necessary derivatives. Let's first express \( z \) in terms of \( t \) through \( x \) and \( y \): Given: - \( z = 3x \sin y \) - \( x = t^4 \) - \( y = 3t^2 \) Now, find the partial derivatives: 1. Calculate \( \frac{dz}{dx} \) and \( \frac{dz}{dy} \): \[ \frac{dz}{dx} = 3 \sin y \] \[ \frac{dz}{dy} = 3x \cos y \] 2. Calculate \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \): \[ \frac{dx}{dt} = 4t^3 \] \[ \frac{dy}{dt} = 6t \] Now, apply the Chain Rule: \[ \frac{dz}{dt} = \frac{dz}{dx} \cdot \frac{dx}{dt} + \frac{dz}{dy} \cdot \frac{dy}{dt} \] Substituting the derivatives we found: \[ \frac{dz}{dt} = (3 \sin(3t^2)) \cdot (4t^3) + (3t^4) \cdot (6t) \cos(3t^2) \] Thus, combining these gives us: \[ \frac{dz}{dt} = 12t^3 \sin(3t^2) + 18t^5 \cos(3t^2) \] So, the final expression is: \[ \frac{dz}{dt} = 12t^3 \sin(3t^2) + 18t^5 \cos(3t^2) \]

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