Question

Find an equation for dy/dx if e^(cos(y))=x^3 arctan (y).

Answer 1

which expression left or right gives you trouble when differentiating ?

Answer 2

The right side

Answer 3

We will have to use the product rule for the right hand expression.
\[\frac{d}{dx}(x^3 \arctan(y))= \arctan(y) \cdot \frac{d(x^3)}{dx}+x^3 \cdot \frac{d(\arctan(y)}{dx}\]

Answer 4

I bet you can find the derivative of x^3 w.r.t x

Answer 5

maybe it is the arctan(y) that is giving you troubles

Answer 6

do you recall this
\[\frac{d}{dx}\arctan(x)=\frac{1}{1+x^2} \]?

Answer 7

if you can recall that and chain rule then you would know that
\[\frac{d}{dx} \arctan(u)=\frac{1}{1+u^2} \cdot \frac{du}{dx} \text{ where } u=u(x)\]

Answer 8

Whenever you come across a problem, try inverting it, for instance a simple change like this makes the problem easier to solve without really changing anything. \[\large y=\arctan x \\ \large \tan y = x\]

Answer 9

Yes it rings a bell. So all of it put together would it be
\[-y'\sin(y) e^\cos (y) = 3x^2 \arctan (y) + x^3 (1/1+y^2)y'\]

Answer 10

that would make it look uglier wouldn't it

if that x^3 thing wasn't there didn't it might make it prettier

Answer 11

yes from what i can tell it looks good

Answer 12

\[-y' \sin(y) e^{\cos(y)}=3x^2 \arctan(y)+x^3 \frac{1}{1+y^2} y'\]

now you have to isolate your y'
do you know how to do this part?

Answer 13

This is where I get stuck

Answer 14

identify which terms have the factor y' (we we call these the y' terms )
and
identify the ones without the factor y' ( we will call these the non-y' terms)

get your y' terms on one side and your non y' terms on the opposite
do these by adding and subtracting terms on both sides

Answer 15

this*

Answer 16

For example.
\[f(x,y)y'=g(x,y)+h(x,y)y' \\ \text{ solve for } y' \\ f(x,y)y'-h(x,y)y'=g(x,y) \\ (f(x,y)-h(x,y))y'=g(x,y) \\ y'=\frac{g(x,y)}{f(x,y)-h(x,y)}\]

Answer 17

In that first step I subtracted h(x,y)y' on both sides

Answer 18

I believe I have found y'. Would that be the equation. I am going to post it to see if it is correct.

Answer 19

\[y'=\frac{ 3x^2\arctan(y) }{ \sin(y)e ^{\cos (y)}-\frac{ x^3 }{ 1+y^2 }}\]

Answer 20

beautiful but i think there is one very small type-o

Answer 21

\[y'=\frac{3x^2 \arctan(y)}{-\sin(y)e^{\cos(y)}-\frac{x^3}{1+y^2}} \]

Answer 22

you could get rid of the compound fraction by multiplying top and bottom by (1+y^2)

Answer 23

\[y'=\frac{3 x^2 \arctan(y)(1+y^2)}{-\sin(y)e^{\cos(y)}(1+y^2)-x^3}\]

Answer 24

Okay I forgot the negative sign. Would this be the equation for dy/dx?

Answer 25

yeah

Answer 26

Thank you. There are two problems I am stuck with. Can you help me through those here too.

Answer 27

i have to go
i'm sorry
bed time awaits