Partial Derivative wrt z of z^2 equals x^2 - 2 x y - 1 at (1, -2, -2)/Implicit Method

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.

$\Box$

\(\ds z^2\)

\(=\)

\(\ds x^2 - 2 x y - 1\)

\(\ds 2 z \dfrac {\partial z} {\partial x}\)

\(=\)

\(\ds 2 x - 2 y\)

Power Rule for Derivatives, Chain Rule for Derivatives

\(\ds \leadsto \ \ \)

\(\ds \dfrac {\partial z} {\partial x}\)

\(=\)

\(\ds \frac {2 \paren 1 - 2 \paren {-2} } {2 \paren {-2} }\)

substituting $x = 1$, $y = -2$ and $z = -2$

\(\ds \)

\(=\)

\(\ds \frac 6 {-4}\)

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 $4$