**Problem Statement:**Find \(\frac{dy}{dt}\).Given:\[ y = \cot^2 \left( \cos^5 t \right) \]**Solution:**To find \(\frac{dy}{dt}\), apply the chain rule and the derivative formulas for trigonometric functions. Remember, the derivative of \(\cot(u)\) with respect to \(u\) is \(-\csc^2(u)\), and you need to apply the chain rule multiple times.**Step-by-Step Solution:**1. Differentiate \( y = \cot^2(u) \) with respect to \(u\): \[ \frac{dy}{du} = 2 \cot(u) \cdot (-\csc^2(u)) \]2. Substitute \( u = \cos^5(t) \) back into the equation: \[ \frac{dy}{du} = -2 \cot(\cos^5(t)) \cdot \csc^2(\cos^5(t)) \]3. Differentiate \( u = \cos^5(t) \) with respect to \(t\): \[ \frac{du}{dt} = 5 \cos^4(t)(-\sin(t)) = -5 \cos^4(t) \sin(t) \]4. Apply the chain rule: \[ \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} \]Combine all the derived expressions to get:\[\frac{dy}{dt} = -2 \cot(\cos^5(t)) \cdot \csc^2(\cos^5(t)) \cdot (-5 \cos^4(t) \sin(t))\]This results in:\[\frac{dy}{dt} = 10 \cot(\cos^5(t)) \cdot \csc^2(\cos^5(t)) \cdot \cos^4(t) \sin(t)\]The solution highlights the application of the chain rule in finding the derivative of trigonometric and power functions, requiring multiple chain rule applications due to the composition of functions.

Name: **Problem Statement:**Find \(\frac{dy}{dt}\).Given:\[ y = \cot^2 \lef
Uploaded: 2024-03-20T06:46:42.000Z
Channel: Bartleby
Description: **Problem Statement:**Find \(\frac{dy}{dt}\).Given:\[ y = \cot^2 \left( \cos^5 t \right) \]**Solution:**To find \(\frac{dy}{dt}\), apply the chain rule and the derivative formulas for trigonometric functions. Remember, the derivative of \(\cot(u)\)