Question

If log (x.y)=x^2+y^2 , find dy/dx?

Answer 1

Please see below.

Explanation:

Here we have to use implicit differentiation.

As `log(xy)=x^2+y^2` or `logx+logy=x^2+y^2`

assuming base is `e`, we have on differentiation

`1/x+1/y(dy)/(dx)=2x+2y(dy)/(dx)`

or `(dy)/(dx)(2y-1/y)=1/x-2x`

or `(dy)/(dx)=(1/x-2x)/(2y-1/y)=(y-2x^2y)/(2xy^2-x)`

In case base is `10`, we can write it as

`lnx/ln10+lny/ln10=x^2+y^2`

and differentiating gives us

`1/ln10(1/x+1/y(dy)/(dx))=2x+2y(dy)/(dx)`

or `(dy)/(dx)(2yln10-1/y)=-2xln10+1/x`

or `(dy)/(dx)=(-2xln10+1/x)/(2yln10-1/y)=(y-2x^2yln10)/(2xy^2ln10-x)`