Find each Derivative \( 41 y=\left(\frac{3 t^{2}-1}{3 t^{2}+1}\right)^{-3} \)
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To find the derivative of \( 41 y=\left(\frac{3 t^{2}-1}{3 t^{2}+1}\right)^{-3} \), we start by applying the chain rule and the quotient rule. First, let \( u = \frac{3 t^2 - 1}{3 t^2 + 1} \). Then, we need to find \( \frac{dy}{dt} \) as follows: Using the chain rule: 1. Differentiate \( y \) with respect to \( u \): \[ \frac{dy}{du} = 41 \cdot (-3) u^{-4} = -123 u^{-4} \] 2. Now, we differentiate \( u \) with respect to \( t \) using the quotient rule: \[ \frac{du}{dt} = \frac{(3 t^2 + 1)(6t) - (3 t^2 - 1)(6t)}{(3 t^2 + 1)^2} \] Simplifying this gives: \[ \frac{du}{dt} = \frac{6t[(3t^2 + 1) - (3t^2 - 1)]}{(3t^2 + 1)^2} = \frac{6t \cdot 2}{(3t^2 + 1)^2} = \frac{12t}{(3t^2 + 1)^2} \] Now, combine these results to find the derivative \( \frac{dy}{dt} \): \[ \frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} = -123 u^{-4} \cdot \frac{12t}{(3t^2 + 1)^2} \] Substituting \( u \) back in: \[ \frac{dy}{dt} = -123 \left(\frac{3t^2 - 1}{3t^2 + 1}\right)^{-4} \cdot \frac{12t}{(3t^2 + 1)^2} \] This will give the final derivative, and can be simplified further if necessary.
