Partial Derivative wrt z of z^2 equals x^2 - 2 x y - 1 at (1, -2, -2)/Explicit Method
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Theorem
Let $z^2 = x^2 - 2 x y - 1$.
Then:
- $\valueat {\dfrac {\partial z} {\partial x} } {x \mathop = 1, y \mathop = -2, z \mathop = -2} = -\dfrac 3 2$
Proof
First we make sure that $\tuple {1, -2, -2}$ actually satisfies the equation:
| \(\ds x^2 - 2 x y - 1\) | \(=\) | \(\ds \paren 1^2 - 2 \paren 1 \paren {-2} - 1\) | at $\tuple {1, -2, -2}$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds 1 + 4 - 1\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 4\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {-2}^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds z^2\) | at $\tuple {1, -2, -2}$ |
and it is seen that this is the case.
$\Box$
| \(\ds z^2\) | \(=\) | \(\ds x^2 - 2 x y - 1\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds z\) | \(=\) | \(\ds \pm \sqrt {x^2 - 2 x y - 1}\) |
When $z = -2$ we have that $\sqrt {x^2 - 2 x y - 1} < 0$.
Hence:
- $z = -\sqrt {x^2 - 2 x y - 1}$
Thus we continue:
| \(\ds \leadsto \ \ \) | \(\ds \dfrac {\partial z} {\partial x}\) | \(=\) | \(\ds -\frac 1 {2 \sqrt {x^2 - 2 x y - 1} } \cdot \paren {2 x - 2 y}\) | Power Rule for Derivatives, Chain Rule for Derivatives | ||||||||||
| \(\ds \) | \(=\) | \(\ds -\frac {2 x - 2 y} {2 \sqrt {x^2 - 2 x y - 1} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds -\frac {2 \paren 1 - 2 \paren {-2} } {2 \sqrt {\paren 1^2 - 2 \paren 1 \paren {-2} - 1} }\) | substituting $x = 1$ and $y = -2$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\frac 6 {2 \sqrt 4}\) | simplifying | |||||||||||
| \(\ds \) | \(=\) | \(\ds -\frac 3 2\) | simplifying |
$\blacksquare$
Sources
- 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $4$