Total Diff: Calculate d(x*y^4)

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To calculate the total differential d(x * y^4) when x = 2, y = 3, dx = 0.02, and dy = -0.03, the product rule is applied, yielding d(xy^4) = y^4 dx + 4xy^3 dy. Substituting the values results in d(xy^4) = (3^4)(0.02) + 4(2)(3^3)(-0.03), which simplifies to -4.86. A key discussion point clarifies that when differentiating with respect to y, x is treated as a constant, leading to the term 4xy^3 dy rather than 4y^3 dy. The conversation also touches on the importance of accurately interpreting the question and the difference between exact calculations and approximations using differential coefficients.
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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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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.
 
Oh I thought you had to remove x . Now I understand thank you
 
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.
 
Nanu Nana said:
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.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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