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## Homework Statement

Calculate d (x * y ^ 4) if x = 2, y = 3, dx and dy = 0.02 = -0.03

## Homework Equations

Total differential

## The Attempt at a Solution

product rule:

d(xy^4) = d/dx (xy^4) dx + d/dy (xy^4) dy

d(xy^4) = y^4 dx + 4xy^3 dy

When x = 2, y = 3, dx = 0.02, y = −0.03

d(xy^4) = (3)^4 (0.02) + 4(2)(3)^3 (−0.03) = 1.62−6.48 = −4.86

Why is d/dy(xy^4)dy not = 4y^3dy but its 4xy^3dy ?

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