Total differential

[Nanu Nana]

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 ?

 

[Guneykan Ozgul]

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.

 

[Nanu Nana]

Oh I thought you had to remove x . Now I understand thank you

 

[HallsofIvy]

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.

 

[epenguin]

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.