Question

I want y' of this function:

Answer 1

\[-\sin(x ^{2}+2y).(2x+2y \prime)=x.e ^{y ^{2}}.2y.y \prime+e ^{y ^{2}}\]

Answer 2

Isolating y' should be easy, apply distributive property in LHS

Answer 3

After this, get all term with y' to one side and all term without y' to other, then you can factor y' out

Answer 4

Can you do this? @soso707

Answer 5

\[2y \prime.y \prime=x.e ^{y ^{2}}.2y+e ^{y ^{2}}-2x+\sin(x ^{2}+2y)\]
it's true?

Answer 6

:(

Answer 7

help me :(

Answer 8

\(-\sin(x ^{2}+2y).(2x+2y \prime)=x.e ^{y ^{2}}.2y.y \prime+e ^{y ^{2}}\ \\
-2x \sin (x^{2} + 2y) -2y' (\sin (x^{2} + 2y) ) = x.e ^{y ^{2}}.2y.y \prime+e ^{y ^{2}} \)

\(2y' (\sin (x^2 + 2y)) + x.e^{y^2}. 2y.y' = -2x \sin (x^2 +2y) - e^{y ^2} \)

\(y' ( 2\sin (x^2 +2y) + x .e^{y^2} . 2y ) = -2x \sin (x^2 +2y) -e^{y^2} \)

Answer 9

\(y' = \cfrac{-2x \sin (x^2 +2y) - e^{y^2} }{2\sin (x^2 +2y) + x.e^{y^2}. 2y} \)

This is not so easy.. I think. Wait, I will try something new.

Answer 10

\[-\sin(x ^{2}+2y).(2x+2y \prime)=x.e ^{y ^{2}}.2y.y \prime+e ^{y ^{2}}\ \\ -2x \sin (x^{2} + 2y) -2y' (\sin (x^{2} + 2y) ) = x.e ^{y ^{2}}.2y.y \prime+e ^{y ^{2}} \]
how can you get this part?

Answer 11

Using distributive property :

a(b+c) = ab + ac

\((2x + 2y')((-\sin (x^2 +2y)) = 2x \times \left( - \sin (x^2 + 2y) \right) + 2y' \times \left( - \sin (x^2 +2y) \right) \)

Answer 12

aha

Answer 13

so the fainl solution

\[y' = \cfrac{-2x \sin (x^2 +2y) - e^{y^2} }{2\sin (x^2 +2y) + x.e^{y^2}. 2y} \]

Answer 14

So, have you got it?

Answer 15

yes from you

Answer 16

thank you

Answer 17

You're Welcome Soso :)