$\frac{1}{2.03}\+\frac{1}{3.02}\+\frac{1}{4.97}$

$\doteq f\left(2\,3\,5\right)\+\mathrm{df}$

$\=\frac{1}{2}\+\frac{1}{3}\+\frac{1}{5}\+\left(f_x\left(2\,3\,5\right)\cdot \left(.03\right)\+f_y\left(2\,3\,5\right)\cdot \left(.02\right)\+f_z\left(2\,3\,5\right)\cdot \left(-\mathrm{.}03\right)\right)$

$\=\frac{31}{30}\+\left(-\frac{1}{4}\cdot \left(.03\right)-\frac{1}{9}\cdot \left(.02\right)-\frac{1}{25}\cdot \left(-.03\right)\right)$

$\=1.033333333-0.008522222222$

$\=1.024811111$

Define $f\left(x\,y\right)$ and its first partial derivatives

$f\left(x\,y\,z\right)\=\frac{1}{x}\+\frac{1}{y}\+\frac{1}{z}$$\stackrel{\text{assign as function}}{\to }$$f$

Calculate $f\left(2.03\,3.02\,4.97\right)\doteq f\left(2\,3\,5\right)\+\mathrm{df}$ 

$f\left(2\,3\,5\right)\+\left(\mathrm{f\_\_x}\left(2\,3\,5\right)\cdot \left(.03\right)\+\mathrm{f\_\_y}\left(2\,3\,5\right)\cdot \left(.02\right)\+\mathrm{f\_\_z}\left(2\,3\,5\right)\cdot \left(-.03\right)\right)$ = $1.024811111$

Compute $f\left(2.03\,3.02\,4.97\right)$ "exactly"

$f\left(2.03\,3.02\,4.97\right)$ = $1.024943909$$$

Define an appropriate function $f$ 

$f≔\left(x\,y\,z\right)\to 1\/x\+1\/y\+1\/z\:$

Compute $\mathrm{df}\=f_x\left(2\,3\,5\right)\cdot \left(0.03\right)\+f_y\left(2\,3\,5\right)\cdot \left(0.02\right)\+f_z\left(2\,3\,5\right)\cdot \left(-0.03\right)$ 

$\mathrm{df}≔{\mathrm{D}}_1\left(f\right)\left(2\,3\,5\right)\cdot \left(.03\right)\+{\mathrm{D}}_2\left(f\right)\left(2\,3\,5\right)\cdot \left(.02\right)\+{\mathrm{D}}_3\left(f\right)\left(2\,3\,5\right)\cdot \left(-.03\right)$

$-0.008522222222$

Approximate $f\left(2.03\,3.02\,4.97\right)$ as $f\left(2\,3\,5\right)\+\mathrm{df}$ 

$\mathrm{evalf}\left(f\left(2\,3\,5\right)\+\mathrm{df}\right)$ = $1.024811111$$$

Obtain $f\left(2.03\,3.02\,4.97\right)$ "exactly"

$f\left(2.03\,3.02\,4.97\right)$ = $1.024943909$$$