Answer
a) $∂w/∂x|_{(2,1,-1)} = \frac{2}{\sqrt{7}}$
b) $∂w/∂y|_{(2,1,-1)} = \frac{4}{\sqrt{7}}$
c) $∂w/∂z|_{(2,1,-1)} = \frac{1}{\sqrt{7}}$
Work Step by Step
Let w = $\sqrt{x^{2}+4y^{2}-z^{2}}$
a) Calculate the partial derivative $\frac{∂w}{∂x}$
$\frac{∂w}{∂x} = \frac{∂}{∂x}[\sqrt{x^{2}+4y^{2}-z^{2}}]$
$\frac{∂w}{∂x} = \frac{1}{2\sqrt{x^{2}+4y^{2}-z^{2}}} \frac{∂}{∂x}[x^{2}+4y^{2}-z^{2}]$
$\frac{∂w}{∂x} = \frac{1}{2\sqrt{x^{2}+4y^{2}-z^{2}}}2x$
$\frac{∂w}{∂x} = \frac{x}{\sqrt{x^{2}+4y^{2}-z^{2}}}$
Calculate the partial derivative $\frac{∂w}{∂x}$ at point (2,1,-1)
$\frac{∂w}{∂x} = \frac{x}{\sqrt{x^{2}+4y^{2}-z^{2}}}$
$∂w/∂x|_{(2,1,-1)} = \frac{2}{\sqrt{2^{2}+4(1)^{2}-(-1)^{2}}}$
$∂w/∂x|_{(2,1,-1)} = \frac{2}{\sqrt{7}}$
b) Calculate the partial derivative $\frac{∂w}{∂y}$
$\frac{∂w}{∂y} = \frac{∂}{∂y}[\sqrt{x^{2}+4y^{2}-z^{2}}]$
$\frac{∂w}{∂y} = \frac{1}{2\sqrt{x^{2}+4y^{2}-z^{2}}} \frac{∂}{∂y}[x^{2}+4y^{2}-z^{2}]$
$\frac{∂w}{∂y} = \frac{1}{2\sqrt{x^{2}+4y^{2}-z^{2}}}8y$
$\frac{∂w}{∂y} = \frac{4y}{\sqrt{x^{2}+4y^{2}-z^{2}}}$
Calculate the partial derivative $\frac{∂w}{∂y}$ at point (2,1,-1)
$\frac{∂w}{∂y} = \frac{4y}{\sqrt{x^{2}+4y^{2}-z^{2}}}$
$∂w/∂y|_{(2,1,-1)} = \frac{4(1)}{\sqrt{2^{2}+4(1)^{2}-(-1)^{2}}}$
$∂w/∂y|_{(2,1,-1)} = \frac{4}{\sqrt{7}}$
c) Calculate the partial derivative $\frac{∂w}{∂z}$
$\frac{∂w}{∂z} = \frac{∂}{∂z}[\sqrt{x^{2}+4y^{2}-z^{2}}]$
$\frac{∂w}{∂z} = \frac{1}{2\sqrt{x^{2}+4y^{2}-z^{2}}} \frac{∂}{∂z}[x^{2}+4y^{2}-z^{2}]$
$\frac{∂w}{∂z} = \frac{1}{2\sqrt{x^{2}+4y^{2}-z^{2}}}(-2z)$
$\frac{∂w}{∂z} = \frac{-z}{\sqrt{x^{2}+4y^{2}-z^{2}}}$
Calculate the partial derivative $\frac{∂w}{∂z}$ at point (2,1,-1)
$\frac{∂w}{∂z} = \frac{-z}{\sqrt{x^{2}+4y^{2}-z^{2}}}$
$∂w/∂z|_{(2,1,-1)} = \frac{-(-1)}{\sqrt{2^{2}+4(1)^{2}-(-1)^{2}}}$
$∂w/∂z|_{(2,1,-1)} = \frac{1}{\sqrt{7}}$
