Let's solve each problem step by step, using the Chain Rule to find all partial derivatives. --- ### 1. \( f(x, y) = e^x \tan y + e^{-y} \arctan x \), where \( x = t^2 \) and \( y = 2t \). We are asked to find \(\frac{df}{dt}\) using the chain rule. #### **Step 1: Compute Partial Derivatives** - \(\frac{\partial f}{\partial x} = e^x \tan y + e^{-y} \frac{1}{1 + x^2}\) - \(\frac{\partial f}{\partial y} = e^x \sec^2 y - e^{-y} \arctan x\) #### **Step 2: Compute Derivatives of \(x\) and \(y\) with Respect to \(t\)** - \(\frac{dx}{dt} = 2t\) - \(\frac{dy}{dt} = 2\) #### **Step 3: Chain Rule** \[ \frac{df}{dt} = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} \] Plug in values: \[ \frac{df}{dt} = \left[ e^x \tan y + e^{-y} \frac{1}{1 + x^2} \right] (2t) + \left[ e^x \sec^2 y - e^{-y} \arctan x \right] (2) \] Now substitute \(x = t^2\), \(y = 2t\): \[ \boxed{ \frac{df}{dt} = 2t \left[ e^{t^2} \tan(2t) + e^{-2t} \frac{1}{1 + t^4} \right] + 2 \left[ e^{t^2} \sec^2(2t) - e^{-2t} \arctan(t^2) \right] } \] --- ### 2. \( f(x, y, z) = \frac{1}{\sqrt{x^2 + y^2 + z^2}} \), where \( x = \sin(st), y = \cos(st), z = st^2 \). We are asked to find all partial derivatives using the chain rule. #### **Step 1: Compute Partial Derivatives** Let \( R = \sqrt{x^2 + y^2 + z^2} \): - \(\frac{\partial f}{\partial x} = -\frac{x}{(x^2 + y^2 + z^2)^{3/2}} \) - \(\frac{\partial f}{\partial y} = -\frac{y}{(x^2 + y^2 + z^2)^{3/2}} \) - \(\frac{\partial f}{\partial z} = -\frac{z}{(x^2 + y^2 + z^2)^{3/2}} \) #### **Step 2: Find Partial Derivatives of \(x, y, z\) with Respect to \(s\) and \(t\)** - \(\frac{\partial x}{\partial s} = t \cos(st)\) - \(\frac{\partial y}{\partial s} = -t \sin(st)\) - \(\frac{\partial z}{\partial s} = t^2\) - \(\frac{\partial x}{\partial t} = s \cos(st)\) - \(\frac{\partial y}{\partial t} = -s \sin(st)\) - \(\frac{\partial z}{\partial t} = 2st\) #### **Step 3: Chain Rule** \[ \frac{\partial f}{\partial s} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial s} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial s} + \frac{\partial f}{\partial z} \frac{\partial z}{\partial s} \] \[ \frac{\partial f}{\partial t} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t} + \frac{\partial f}{\partial z} \frac{\partial z}{\partial t} \] #### **Plug in the values:** ##### For \( \frac{\partial f}{\partial s} \): \[ \frac{\partial f}{\partial s} = -\frac{x}{R^3} (t\cos(st)) - \frac{y}{R^3} (-t\sin(st)) - \frac{z}{R^3} (t^2) \] \[ = \frac{1}{R^3} \left[ -x t \cos(st) + y t \sin(st) - z t^2 \right] \] ##### For \( \frac{\partial f}{\partial t} \): \[ \frac{\partial f}{\partial t} = -\frac{x}{R^3} (s\cos(st)) - \frac{y}{R^3} (-s\sin(st)) - \frac{z}{R^3} (2st) \] \[ = \frac{1}{R^3} \left[ -x s \cos(st) + y s \sin(st) - 2z s t \right] \] Now substitute \(x = \sin(st), y = \cos(st), z = st^2, R = \sqrt{\sin^2(st) + \cos^2(st) + (st^2)^2} = \sqrt{1 + s^2 t^4}\): #### **Final Answers:** \[ \boxed{ \frac{\partial f}{\partial s} = \frac{1}{\left(1 + s^2 t^4\right)^{3/2}} \left[ -\sin(st) t \cos(st) + \cos(st) t \sin(st) - s t^4 \right] } \] \[ \boxed{ \frac{\partial f}{\partial t} = \frac{1}{\left(1 + s^2 t^4\right)^{3/2}} \left[ -\sin(st) s \cos(st) + \cos(st) s \sin(st) - 2 s^2 t^3 \right] } \] --- If you'd like, I can further simplify the trigonometric expressions! Let me know if you need that.

1. \(\displaystyle \frac{df}{dt} = 2t\left(e^{t^2}\tan 2t + \frac{e^{-2t}}{1 + t^4}\right) + 2\left(e^{t^2}\sec^2 2t - e^{-2t}\arctan t^2\right)\) 2. \(\displaystyle \frac{\partial f}{\partial s} = \frac{-\sin(st) t \cos(st) + \cos(st) t \sin(st) - s t^4}{\left(1 + s^2 t^4\right)^{3/2}}\) 3. \(\displaystyle \frac{\partial f}{\partial t} = \frac{-\sin(st) s \cos(st) + \cos(st) s \sin(st) - 2 s^2 t^3}{\left(1 + s^2 t^4\right)^{3/2}}\)