how can i find the approximate value of this?

Question

Using the total differential, find the approximate value of:

$$8.02^{(\frac{1}{3})}*\sqrt{(3.96)^2+(3.03)^2}$$

how can I find the approximate value of this using partial derivatives, explain, please

Answer 1

assume $$f(x,y,z)=(x)^{\frac{1}{3}}*\sqrt{y^2+z^2}$$ $$x=8;y=4;z=3$$you get: $$f(x,y,z)=10$$use this: $$df=\frac{\partial{f}}{\partial{x}}dx+\frac{\partial{f}}{\partial{y}}dy+\frac{\partial{f}}{\partial{z}}dz$$

$$dx=0.02;dy=-0.04;dz=0.03$$

find the partials and solve, then find $f(8,4,3)+df$, you must get around$(9.98)$

Answer 2

Define $u{(x,y,z)} = x^{1/3}\sqrt{y^2+z^2}$

Assume $x=8$, $y=4$, $z=3$

$dx = 0.02$, $dy = -0.04$ and $dz = +0.03$

Then find

$u(8,4,3)$

and

$du = \frac{\partial u}{\partial x}dx + \frac{\partial u}{\partial }dy+\frac{\partial u}{\partial z}dz$

The approximate value will be

$u(8,4,3)+du$