differentiation 4

[red and white kop!]

(x^2)(y^2)= (y+1)/(x+1)
find dy/dx
i differentiated to finally get 2xy(y + x (dy/dx)) (x+1)^2 = dy/dx (x+1) - (y+1)
but from here im not sure how to extract those two dy/dx

 

[Deleted member 4993]

(x^2)(y^2)= (y+1)/(x+1)
find dy/dx
i differentiated to finally get 2xy(y + x (dy/dx)) (x+1)^2 = dy/dx (x+1) - (y+1)
but from here im not sure how to extract those two dy/dx

By sixth grade algebra!

2xy[sup:1gbsphi0]2[/sup:1gbsphi0](x+1)[sup:1gbsphi0]2[/sup:1gbsphi0] + 2x[sup:1gbsphi0]2[/sup:1gbsphi0]y(x+1)[sup:1gbsphi0]2[/sup:1gbsphi0] dy/dx = (x+1)dy/dx - (y+1)

[2x[sup:1gbsphi0]2[/sup:1gbsphi0]y(x+1) - 1](x+1) dy/dx = - (y+1) - 2xy[sup:1gbsphi0]2[/sup:1gbsphi0](x+1)[sup:1gbsphi0]2[/sup:1gbsphi0]

and it is done....

 

[red and white kop!]

hm
thats not the answer in the book

 

[BigGlenntheHeavy]

\(\displaystyle x^{2}y^{2} \ = \ \frac{y+1}{x+1}, \ (x^{2}y^{2})(x+1) \ = \ (y+1)\)

\(\displaystyle x^{3}y^{2}+x^{2}y^{2} \ = \ y+1, \ 3x^{2}y^{2}+2x^{3}yy'+2xy^{2}+2x^{2}yy' \ = \ y'\)

\(\displaystyle 2x^{3}yy'+2x^{2}yy'-y' \ = \ -3x^{2}y^{2}-2xy^{2}\)

\(\displaystyle y' \ = \ \frac{-3x^{2}y^{2}-2xy^{2}}{2x^{3}y+2x^{2}y-1} \ = \ -\frac{3x^{2}y^{2}+2xy^{2}}{2x^{3}y+2x^{2}y-1}\)

 

[red and white kop!]

can this be factorised? i got a similar expression but i was unable to factorise it

 

[BigGlenntheHeavy]

Well, what is the answer in the book!

 

[red and white kop!]

the answer is -[y(y+1)(3x+2)]/[x(x+1)(y+2)]
i got the same answer as you but i cant factorise it this way :x

 

[BigGlenntheHeavy]

A bunch of needless grunt work. If you want the answer in the book, sub in x^2y^2, otherwise leave your final answer alone.

 

[red and white kop!]

thanks!