The partial differential expression is equal to the expression 30 · x · (x³ + y⁴) + 36 · x² · y² · (y² + x²) and the total differential formula is Δz = (0.9 · x² + 0.6 · y²) · cos (x³ + y³).
How to apply partial derivatives and total differentials
In the first part of this question we must apply the concept of partial derivatives to find the form of the entire expression, whose variables are two. Partial differentiation is a generalization of the ordinary differentiation:
[tex]\frac{\partial f}{\partial x} = 6\cdot x^{5}+12\cdot x^{3}\cdot y^{4}+y^{6}[/tex] (1)
[tex]\frac{\partial^{2} f}{\partial x^{2}} = 30\cdot x^{4}+36\cdot x^{2}\cdot y^{4}[/tex] (2)
[tex]\frac{\partial^{2} f}{\partial x \,\partial y} = 48\cdot x^{3}\cdot y^{3}+6\cdot y^{5}[/tex] (3)
[tex]\frac{\partial f}{\partial y} = 12\cdot x^{4}\cdot y^{3} +6\cdot x \cdot y^{5}[/tex] (4)
[tex]\frac{\partial^{2} f}{\partial y^{2}} = 36\cdot x^{4}\cdot y^{2} + 30\cdot x\cdot y^{4}[/tex] (5)
[tex]\frac{\partial f}{\partial y \,\partial x} = 48\cdot x^{3}\cdot y^{3}+6\cdot y^{5}[/tex] (6)
Then, the entire expression is:
30 · x⁴ + 36 · x² · y⁴ + 36 · x⁴ · y² + 30 · x · y⁴
30 · (x⁴ + x · y⁴) + 36 · (x² · y⁴ + x⁴ · y²)
30 · x · (x³ + y⁴) + 36 · x² · y² · (y² + x²)
In the second part we must determine the total differential of z with respect to two variables:
[tex]\Delta z = \frac{\partial f}{\partial x}\cdot \Delta x + \frac{\partial f}{\partial y}\cdot \Delta y[/tex] (7)
If we know that z = sin (x³ + y³), Δx = 0.3 and Δy = 0.2, then the total differential of the function is:
Δz = 3 · cos (x³ + y³) · x² · (0.3) + 3 · cos (x³ + y³) · y² · (0.2)
Δz = (0.9 · x² + 0.6 · y²) · cos (x³ + y³)
To learn more on partial derivatives: https://brainly.com/question/14782333
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