Question

@hero

Answer 1

find dy/dx given \(sin(x+y)=y^2cosx\)

Answer 2

I did
\[\frac{d}{dx}(sin(x+y))=\frac{d}{dx}(y^2cosx)\\cox(x+y)(1+\frac{dy}{dx})=y^2(-sinx)+2y\frac{dy}{dx}(cosx)\]

Answer 3

Do you have an idea of what to do next?

Answer 4

try to solve for dy/dx but i don't know how

Answer 5

Let's re-write it this way:
\(\cos(x+y)(1+y')=2yy'\cos(x)-y^2(\sin(x)\)

Answer 6

And then like this by associative and commutative properties:
\((y'+1)(\cos(x+y)=2yy'\cos(x)-y^2\sin(x)\)

Answer 7

\((y'+1)(\cos(x+y)=2\cos(x)yy'-y^2\sin(x)\)

Answer 8

a lot of maneuvering

Answer 9

Yes

Answer 10

Factor out the y on the RHS
\((y'+1)(\cos(x+y)=y(2\cos(x)y'-y\sin(x))\)

Answer 11

why the y?

Answer 12

Okay we shouldn't factor out the y but instead to do this
\((y'+1)(\cos(x+y)=2\cos(x)yy'-y^2\sin(x)\)

Apply the distributive property on the LHS:
\(\cos(x+y)y' + \cos(x+y) = 2\cos(x)yy'-y^2\sin(x)\)

Answer 13

Actually it's better to do it this way:
Add \(y^2\sin(x)\) to both sides then subtract \(\cos(x+y)y'\) from both sides:

Answer 14

\(\cos(x+y)+y^2\sin(x) = 2\cos(x)yy' -\cos(x + y)y'\)

Answer 15

Then factor out the \(y'\) on the RHS

Answer 16

\(\cos(x+y)+y^2\sin(x) = y'(2\cos(x)y -\cos(x + y))\)

Answer 17

Then isolate \(y'\) as your final step

Answer 18

You should get
\(y' = \dfrac{\cos(x+y)+\sin(x)y^2}{2\cos(x)y-\cos(x+y)}\)

Answer 19

okay, dang
thank you so much!

Answer 20

You're welcome