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

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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.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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