## Problem Statement

Approximately calculate using the differential.

$y=\sqrt{1+x+\sin x}, x=0.01$
## Solution

$y\left(x_{0}+\Delta x\right) \approx y\left(x_{0}\right)^{\prime} \cdot \Delta x+y\left(x_{0}\right)$

$\Delta x=x-x_{0}$

$x=0.01$

$x_{0}=0$

$\Delta x=0.01$

$y\left(x_{0}\right)=y(0)=1$

$y^{\prime}=\frac{1}{2 \cdot \sqrt{1+x+\sin x}} \cdot(1+x+\sin x)^{\prime}=\frac{1+\cos x}{2 \cdot \sqrt{1+x+\sin x}}$

$y^{\prime}\left(x_{0}\right)=y^{\prime}(0)=\frac{1+1}{2 \cdot(1+0+0)}=1$

$\sqrt{1+0.01+\sin 0.01} \approx 1 \cdot 0.01+1=1.01$

## Problem Kuznetsov Differentiation 5-27