Chapter 3 - Section 3.10 - Linear Approximations and Differentials. - 3.10 Exercises - Page 259: 11

(a) $dy = e^{-4x}(1-4x)~dx$ (b) $dy = -\frac{2t^3}{\sqrt{1-t^4}}~dt$

(a) $y = xe^{-4x}$ $\frac{dy}{dx} = e^{-4x}-4xe^{-4x}$ $\frac{dy}{dx} = e^{-4x}(1-4x)$ $dy = e^{-4x}(1-4x)~dx$ (b) $y = \sqrt{1-t^4}$ $\frac{dy}{dt} = \frac{1}{2}(1-t^4)^{-1/2}\cdot -4t^3$ $\frac{dy}{dt} = -\frac{2t^3}{\sqrt{1-t^4}}$ $dy = -\frac{2t^3}{\sqrt{1-t^4}}~dt$