Implicit Derivative F43Ff9

1. **State the problem:** We are given the implicit equation $$-\frac{t^2}{y^2} - \cos(6 y) = -4$$ and need to find the derivative $\frac{dy}{dt}$. 2. **Rewrite the equation:** $$-\frac{t^2}{y^2} - \cos(6 y) = -4$$ can be rearranged as $$-\frac{t^2}{y^2} = -4 + \cos(6 y)$$ but we will differentiate implicitly as is. 3. **Differentiate both sides with respect to $t$: ** Using implicit differentiation, remember $y$ is a function of $t$, so apply the chain rule where needed. $$\frac{d}{dt}\left(-\frac{t^2}{y^2}\right) + \frac{d}{dt}\left(-\cos(6 y)\right) = \frac{d}{dt}(-4)$$ 4. **Differentiate each term:** - For $-\frac{t^2}{y^2}$, use the quotient rule or rewrite as $-t^2 y^{-2}$: $$\frac{d}{dt}(-t^2 y^{-2}) = -\left(2 t y^{-2} + t^2 \cdot (-2) y^{-3} \frac{dy}{dt}\right) = -2 t y^{-2} + 2 t^2 y^{-3} \frac{dy}{dt}$$ - For $-\cos(6 y)$, use chain rule: $$\frac{d}{dt}(-\cos(6 y)) = \sin(6 y) \cdot 6 \frac{dy}{dt} = 6 \sin(6 y) \frac{dy}{dt}$$ - The derivative of $-4$ is $0$. 5. **Put it all together:** $$-2 t y^{-2} + 2 t^2 y^{-3} \frac{dy}{dt} + 6 \sin(6 y) \frac{dy}{dt} = 0$$ 6. **Group terms with $\frac{dy}{dt}$:** $$2 t^2 y^{-3} \frac{dy}{dt} + 6 \sin(6 y) \frac{dy}{dt} = 2 t y^{-2}$$ 7. **Factor out $\frac{dy}{dt}$:** $$\left(2 t^2 y^{-3} + 6 \sin(6 y)\right) \frac{dy}{dt} = 2 t y^{-2}$$ 8. **Solve for $\frac{dy}{dt}$:** $$\frac{dy}{dt} = \frac{2 t y^{-2}}{2 t^2 y^{-3} + 6 \sin(6 y)}$$ 9. **Simplify the expression:** Rewrite powers of $y$: $$\frac{dy}{dt} = \frac{2 t / y^{2}}{2 t^{2} / y^{3} + 6 \sin(6 y)} = \frac{2 t y^{-2}}{2 t^{2} y^{-3} + 6 \sin(6 y)}$$ This is the implicit derivative. **Final answer:** $$\boxed{\frac{dy}{dt} = \frac{2 t y^{-2}}{2 t^{2} y^{-3} + 6 \sin(6 y)}}$$