Question

\[f_x(x,y)=\cos x\sin x-xy^2\]\[f_y(x,y)=y(1-x^2)\]find \(f(x,y)\)

Answer 1

how to begin?

Answer 2

Start by integrating f_y with respect to y, you should get\[f(x,y)=\frac{1}{2}(y^2-x^2y^2)+h(x)\]

Answer 3

Then you need to find what h(x) is

Answer 4

\[f(x,y)=\int f_y \text dy=\int y(1-x^2)\text dy=\frac{y^2(1-x^2)}{2}+h(x)\]

Answer 5

Then find the x derivative of f(x,y) you just found, set equal to the original f_x, and solve for h(x)

Answer 6

\[f(x,y)=\int f_x \text dx=\int \cos x\sin x-xy^2\text dx=-\frac {\cos^2x}2+g(y)\]

Answer 7

You just found\[f(x,y)=\frac{1}{2}(y^2-x^2y^2)+h(x)\]Find the x derivative of this so that:\[f_x(x,y)=-xy^2+h'(x)\]We also know that\[f_x(x,y)=\cos x \sin x-xy^2\]So,\[-xy^2+h'(x)=\cos x \sin x -xy^2\]

Answer 8

\[\frac {\partial f(x,y)}{\partial x}=-xy^2+h'(y)\]

Answer 9

ops h'(x)

Answer 10

\[h'(x)=\cos x \sin x\]

Answer 11

Yep, then integrate

Answer 12

\[f(x,y)=-xy^2-\frac{\cos ^2x}2\]

Answer 13

You accidentally put that back into\[f_x(x,y)=-xy^2+h'(x)\] You should have put it back into\[f(x,y)=\frac{1}{2}(y^2-x^2y^2)+h(x)\]

Answer 14

oh

Answer 15

ah yes silly me

Answer 16

\[f(x,y)=\frac{y^2(1-x^2)}{2}-\frac{\cos ^2x}2\]
better?

Answer 17

So you should have 1 of two answers (both the same)\[f(x,y)=\frac{1}{2}(y^2-x^2y^2)-\frac{1}{2}\cos^2 x\]or\[f(x,y)=\frac{1}{2}(y^2-x^2y^2)+\frac{1}{2}\sin^2 x\]

Answer 18

Yes!, except you forgot the minus sign

Answer 19

im not sure i have forgotten the minus sign,

where does the sin^2 term come from?

Answer 20

nvm latex made the minus look like part of the fraction.

The sin comes from the u-substitution if you let u=sinx instead

I checked the solutions in mathematica, we got it right (attachment) :)

Answer 21

im not sure which substitution you mean

Answer 22

\[\int\limits \cos x \sin x dx\]You can let u=sinx or u=-cosx

Answer 23

i dont understand that bit

Answer 24

\[\int\cos x\sin x\text dx\] Let \(u=\sin x\)\[\text du=\cos x \text dx\]\[=\int u\text du=\frac{u^2}2+c\]\[=\sin^2(x)+c\]

Answer 25

\[=\frac{\sin^2(x)}2+c\],
hmm

Answer 26

\[\int\limits \cos x \sin x dx\]Let \[u=\sin x\]\[du=\cos x dx\]
\[\int\limits u du\]\[\frac{1}{2}u^2=\frac{1}{2}\sin^2x\]

or

\[u=\cos x\]\[du=-\sin x\]\[-\int\limits-\sin x \cos x dx\]\[- \int\limits u du=-\frac{1}{2}u^2=-\frac{1}{2}\cos^2 x\]

Answer 27

\[\int\cos x\sin x\text dx\] Let \(u=-\cos x\)\[\text du=\sin x \text dx\]\[=-\int u\text du=-\frac{u^2}2+c\]\[=-\frac{\cos^2(x)}2+c\]

Answer 28

What does the subscript x mean when you write: