Question

Differentiate : xy log(x+y) = 1

Answer 1

Using product rule and chain rule of differentiation:
ylog(x+y) + x(dy/dx)log(x+y) + xy(1+dy/dx)/(x+y) = 0;
Now you can seperate dy/dx on one side using basic mathematics rules.

Answer 2

Given : xy log(x+y) = 1

To find : Differentiate : xy log(x+y) = 1

Solution:

xy log(x+y) = 1

=> log(x + y)  = 1/xy

Differentiating both sides wrt x

=> (1  + dy/dx) / (x  + y )   =     (-1/xy²)dy/dx  -1/x²y

=> 1 + dy/dx=  (x + y ) (-x dy/dx  - y )/(x²y²)

=> x²y² + x²y²dy/dx   =  -(x + y ) (x dy/dx + y)

=> x²y² + x²y²dy/dx  + (x + y ) (x dy/dx + y)  = 0

=> dy/dx  ( x²y² + x² + xy)  +  x²y²  + xy + y² = 0

=> dy/dx = - ( x²y²  + xy + y² ) /  ( x²y² + x² + xy)

dy/dx = - ( x²y²  + xy + y² ) /  ( x²y² + xy + x²)

Learn more:

differentiate wrt x cosx . cos2x . cos3x​ - Brainly.in

https://brainly.in/question/8856618

Differentiate the function w.r.t.x: [tex] m  rac{x an x}{sec x+ an x ...

https://brainly.in/question/7855218