Total differential

  • Thread starter Nanu Nana
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  • #1
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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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Answers and Replies

  • #2
What makes you think that it is 4y^3, what happened to x?
Since the derivation is with respect to y the x will behave like a constant. So it's 4xy^3.
 
  • #3
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Oh I thought you had to remove x . Now I understand thank you
 
  • #4
HallsofIvy
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The derivative of a constant, c, times a function of y, with respect to y. is c times the derivative: d(cf(y))= c (df/dy)dy. And when you are taking the derivative with respect to y, x is treated as a constant.
 
  • #5
epenguin
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Oh I thought you had to remove x . Now I understand thank you
I hope you do understand, not just you 'thought you had to' and now think 'you have to' do something different.

Have you quoted the question exactly? The exact answer to the question as quoted by you is
(2 + 0.02)×(3 - 0.03)4 - 2×3 = ...

Yours is an answer to a question like "use differential coefficients to calculate approximately...".
Compare the result.
 

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