Partial Derivatives of x ln y^2 + y e^z
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Theorem
Let:
- $\map f {x, y, z} = x \ln y^2 + y e^z$
Then:
\(\ds \map {f_1} {1, -1, 0}\) | \(=\) | \(\ds 0\) | ||||||||||||
\(\ds \map {f_2} {x, x y, y + z}\) | \(=\) | \(\ds \dfrac 2 y + e^{y + z}\) |
Proof
\(\ds \map {f_1} {x, y, z}\) | \(=\) | \(\ds \ln y^2\) | Derivative of Constant Multiple, Derivative of Identity Function, holding $y$ and $z$ constant | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {f_1} {1, -1, 0}\) | \(=\) | \(\ds \map \ln {\paren {-1}^2}\) | substituting $\tuple {1, -1, 0}$ for $\tuple {x, y, z}$ | ||||||||||
\(\ds \) | \(=\) | \(\ds \ln 1\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds 0\) | Natural Logarithm of 1 is 0 |
$\Box$
\(\ds \map {f_2} {x, y, z}\) | \(=\) | \(\ds \map {\dfrac \partial {\partial y} } {x \ln y^2 + y e^z}\) | Definition of Partial Derivative | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac x {y^2} \cdot 2 y + e^z\) | Derivative of Natural Logarithm, Power Rule for Derivatives, Chain Rule for Derivatives, holding $x$ and $z$ constant | |||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {2 x} y + e^z\) | simplification | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map {f_2} {x, x y, y + z}\) | \(=\) | \(\ds \dfrac {2 x} {x y} + e^{y + z}\) | substituting $\tuple {x, x y, y + z}$ for $\tuple {x, y, z}$ | ||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac 2 y + e^{y + z}\) | simplifying |
$\blacksquare$
Sources
- 1961: David V. Widder: Advanced Calculus (2nd ed.) ... (previous) ... (next): $1$ Partial Differentiation: $\S 1$. Introduction: Exercise $7$