1. **Problem Statement:** Given the function $u = x^2 + y^2 + z^2$ where $x = e^t$, $y = e^t \cos t$, and $z = e^t \sin t$, find the derivative $\frac{du}{dt}$. 2. **Formula Used:** To find $\frac{du}{dt}$ when $u$ depends on $x$, $y$, and $z$, which in turn depend on $t$, we use the multivariable chain rule: $$\frac{du}{dt} = \frac{\partial u}{\partial x} \frac{dx}{dt} + \frac{\partial u}{\partial y} \frac{dy}{dt} + \frac{\partial u}{\partial z} \frac{dz}{dt}$$ 3. **Partial Derivatives:** Since $u = x^2 + y^2 + z^2$, the partial derivatives are: $$\frac{\partial u}{\partial x} = 2x, \quad \frac{\partial u}{\partial y} = 2y, \quad \frac{\partial u}{\partial z} = 2z$$ 4. **Derivatives of $x$, $y$, and $z$ with respect to $t$:** - $x = e^t \implies \frac{dx}{dt} = e^t$ - $y = e^t \cos t \implies \frac{dy}{dt} = e^t (-\sin t) + \cos t (e^t) = e^t (-\sin t + \cos t)$ - $z = e^t \sin t \implies \frac{dz}{dt} = e^t (\cos t) + \sin t (e^t) = e^t (\cos t + \sin t)$ 5. **Substitute all into the chain rule:** $$\frac{du}{dt} = 2x \cdot e^t + 2y \cdot e^t (-\sin t + \cos t) + 2z \cdot e^t (\cos t + \sin t)$$ 6. **Replace $x$, $y$, and $z$ with their expressions:** $$= 2 e^t \cdot e^t + 2 e^t \cos t \cdot e^t (-\sin t + \cos t) + 2 e^t \sin t \cdot e^t (\cos t + \sin t)$$ 7. **Simplify terms:** $$= 2 e^{2t} + 2 e^{2t} \cos t (-\sin t + \cos t) + 2 e^{2t} \sin t (\cos t + \sin t)$$ 8. **Expand the products:** $$= 2 e^{2t} + 2 e^{2t} (-\sin t \cos t + \cos^2 t) + 2 e^{2t} (\sin t \cos t + \sin^2 t)$$ 9. **Combine like terms:** $$= 2 e^{2t} + 2 e^{2t} (-\sin t \cos t + \cos^2 t + \sin t \cos t + \sin^2 t)$$ 10. **Notice $-\sin t \cos t$ and $+\sin t \cos t$ cancel out:** $$= 2 e^{2t} + 2 e^{2t} (\cos^2 t + \sin^2 t)$$ 11. **Use the Pythagorean identity $\cos^2 t + \sin^2 t = 1$:** $$= 2 e^{2t} + 2 e^{2t} (1) = 4 e^{2t}$$ **Final answer:** $$\frac{du}{dt} = 4 e^{2t}$$