Partial Derivative/Examples/u^2 + v^2 = x^2, 2 u v = 2 x y + y^2/Explicit Method

Example of Partial Derivative

Consider the simultaneous equations:

\(\ds u^2 + v^2\)

\(=\)

\(\ds x^2\)

\(\ds 2 u v\)

\(=\)

\(\ds 2 x y + y^2\)

Then:

$\map {u_1} {1, -2} = 1$

at $u = 1$, $v = 0$.

Proof

By definition of partial derivative:

$\map {u_1} {1, -2} = \valueat {\dfrac {\partial u} {\partial x} } {x \mathop = 1, y \mathop = -2}$

hence the motivation for the abbreviated notation on the left hand side.

Lemma

Consider the simultaneous equations:

\(\ds u^2 + v^2\)

\(=\)

\(\ds x^2\)

\(\ds 2 u v\)

\(=\)

\(\ds 2 x y + y^2\)

Then:

$x = 1$, $y = -2$ is a solution at $u = 1$, $v = 0$.

$\Box$

We have:

\(\ds u^2 + v^2\)

\(=\)

\(\ds x^2\)

\(\ds \leadsto \ \ \)

\(\ds u\)

\(=\)

\(\ds \sqrt {x^2 - v^2}\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds \dfrac 1 {2 \paren {\sqrt {x^2 - v^2} } } \paren {2 x - 2 v \dfrac {\partial v} {\partial x} }\)

Power Rule for Derivatives, Chain Rule for Derivatives

\(\text {(1)}: \quad\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds \dfrac {x - v \dfrac {\partial v} {\partial x} } {\sqrt {x^2 - v^2} }\)

simplifying

At this point we investigate the sign of $\sqrt {x^2 - v^2}$ which is needed.

We see that:

\(\ds u\)

\(=\)

\(\ds \sqrt {x^2 - v^2}\)

\(\ds \leadsto \ \ \)

\(\ds 1\)

\(=\)

\(\ds +\sqrt {x^2 - v^2}\)

Hence when we are at the point of plugging in numbers we will need to take the positive square root.

Then we have:

\(\ds 2 u v\)

\(=\)

\(\ds 2 x y + y^2\)

\(\ds \leadsto \ \ \)

\(\ds v\)

\(=\)

\(\ds \dfrac {2 x y + y^2} {2 u}\)

\(\ds \leadsto \ \ \)

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

\(=\)

\(\ds \dfrac {2 u \map {\dfrac \partial {\partial x} } {2 x y + y^2} - \paren {2 x y + y^2} \map {\dfrac \partial {\partial x} } {2 u} } {\paren {2 u}^2}\)

Quotient Rule for Derivatives

\(\ds \)

\(=\)

\(\ds \dfrac {2 u \times 2 x - \paren {2 x y + y^2} 2 \dfrac {\partial u} {\partial x} } {4 u^2}\)

Power Rule for Derivatives, Chain Rule for Derivatives

\(\text {(2)}: \quad\)

\(\ds \)

\(=\)

\(\ds \dfrac {2 u x - \paren {2 x y + y^2} \dfrac {\partial u} {\partial x} } {2 u^2}\)

Power Rule for Derivatives, Chain Rule for Derivatives

Substituting $\dfrac {\partial v} {\partial x}$ from $(2)$ into $(1)$ gives:

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

\(=\)

\(\ds \dfrac {x - v \paren {\dfrac {2 u x - \paren {2 x y + y^2} \dfrac {\partial u} {\partial x} } {2 u^2} } } {\sqrt {x^2 - v^2} }\)

Could rearrange and simplify, but easier to substitute the values of $x$ and $y$ at this point:

\(\ds \)

\(=\)

\(\ds \dfrac {1 - 0 \times \paren {\cdots} } {+\sqrt {1^2 - 0^2} }\)

$v$ vanishes, hence the big parenthesis is irrelevant

\(\ds \)

\(=\)

\(\ds 1\)

simplifying

$\blacksquare$

Sources

• 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $15$