Find the total derivative of ##u## with respect to ##x##

[chwala]

Homework Statement

see attached

Relevant Equations

total derivatives

see attached below; the textbook i have has many errors...

clearly $f_x$ is wrong messing up the whole working to solution...we ought to have;
$\frac {du}{dx}=(9x^2+2y)+(2x+8y)3=9x^2+2y+6x+24y=9x^2+6x+26y$

 

[anuttarasammyak]

$$\frac{du}{dx}=9x^2+2y+2x\frac{dy}{dx}+8y\frac{dy}{dx}=9x^2+2(3x+5)+2x3+8(3x+5)3$$
Expression $u_x$ seems ambiguous to me. How can we change x without changing y?

 

[Mark44]

Homework Statement:: see attached
Relevant Equations:: total derivatives

see attached below; the textbook i have has many errors...

View attachment 296761

clearly $f_x$ is wrong messing up the whole working to solution...we ought to have;
$u_x=(9x^2+2y)+(2x+8y)3=9x^2+2y+6x+24y=9x^2+6x+26y$

Their answer has a typo. The very first term on the right side should be $9x^2$, not $6x^2$.

 

[chwala]

$$\frac{du}{dx}=9x^2+2y+2x\frac{dy}{dx}+8y\frac{dy}{dx}=9x^2+2(3x+5)+2x3+8(3x+5)3$$
Expression $u_x$ seems ambiguous to me. How can we change x without changing y?

yeah..let me amend that... we want total derivative and not partial derivative...i posted this using my android phone and looks like i did not get it right...