Computing this constant by differential Please compute $((3.8)^2+2(2.1)^3)^{1/5}$ approximately using differentials. Thank you very much.
 A: Note that 3.8 is near 4 and 2.1 is near 2.  Our first approximation, then, without differentials is
$$ (4^2 + 2(2)^3)^{1/5} = 2 $$
Now, we need to approximate the change caused when moving the 4 down to 3.8 and the 2 up to 2.1.  Therefore, we consider the function
$$ z = (x^2 + 2y^3)^{1/5} $$
The total differential of z at $(x_0,y_0)$ is given by
$$ dz = \left.\frac{\partial z}{\partial x}\right|_{x_0,y_0} \Delta x + \left.\frac{\partial z}{\partial y}\right|_{x_0,y_0} \Delta y $$
Think of it this way: we are calculating the tangent plane to the graph of $z$ at an easily computable point $ (x_0,y_0) = (4,2) $.  We want the value of $z$ at $(3.8,2.1)$, but that's hard to compute.  So we assume $z$ is well approximated by its tangent plane, and compute the value based on that.  The total differential is telling us the difference in height of that tangent plane when we move our $x$ and $y$.
In any case, we have $\Delta x = 3.8-4 = -0.2$ and $\Delta y = 2.1-2 = 0.1$.  Also,
$$ \frac{\partial z}{\partial x} = \frac{1}{5}(x^2+2y^3)^{-4/5}(2x) $$
Evaluating this at $(4,2)$:
$$ \frac{\partial z}{\partial x} = \frac{1}{10} $$
Similarly, 
$$ \frac{\partial z}{\partial y} = \frac{1}{5}(x^2+2y^3)^{-4/5}(6y^2) $$
Evaluate at $(4,2)$:
$$ \frac{\partial z}{\partial y} = \frac{3}{10} $$
So, we have 
$$ dz = \frac{1}{10} \cdot (-0.2) + \frac{3}{10} \cdot 0.1 = 0.01 $$
Therefore, the approximation which takes into account differentials is $z + dz = 2.01$
