Chapter 3 - Section 3.10 - Linear Approximations and Differentials. - 3.10 Exercises - Page 256: 17

(a) $dy = \frac{x}{\sqrt{3+x^2}}~dx$ (b) $dy = -0.05$

(a) $y = \sqrt{3+x^2}$ $\frac{dy}{dx} = \frac{1}{2}(3+x^2)^{-1/2}\cdot (2x)$ $\frac{dy}{dx} = \frac{x}{\sqrt{3+x^2}}$ $dy = \frac{x}{\sqrt{3+x^2}}~dx$ (b) $dy = \frac{x}{\sqrt{3+x^2}}~dx$ $dy = \frac{1}{\sqrt{3+(1)^2}}~(-0.1)$ $dy = \frac{1}{2}~(-0.1)$ $dy = -0.05$