I have now shown you how to do several of these kind of derivatives.
It is time for you to show us some of your own efforts.
What have you got so far ?
Differentiate the following implicit dunction and find dy/dx ln (3xy)=2xy^2
4 answers
ln (3x) + ln (y) = 2xy^2
Is it need to separate in the bracket?
Is it need to separate in the bracket?
How to separate x and y for RHS
There is really no need to have brackets on the RS
but I would mentally look at it as
(2x)(y^2) so the product rule is more obvious
so .....
3/(3x) + (dy/dx)/y = (2x)(2y dy/dx) + (y^2)(2)
multiply each term by y
y/x + dy/dx = 4xy^2 dy/dx + 2y^3
dy/dx - 4xy^2 dy/dx = 2y^3 - y/x
dy/dx (1 - 4xy^2) = 2y^3 - y/x
dy/dx = (2y^3 - y/x)/(1 - 4xy^2)
= (2xy^3 - y)/(x - 4x^2 y^2)
again , check my steps
but I would mentally look at it as
(2x)(y^2) so the product rule is more obvious
so .....
3/(3x) + (dy/dx)/y = (2x)(2y dy/dx) + (y^2)(2)
multiply each term by y
y/x + dy/dx = 4xy^2 dy/dx + 2y^3
dy/dx - 4xy^2 dy/dx = 2y^3 - y/x
dy/dx (1 - 4xy^2) = 2y^3 - y/x
dy/dx = (2y^3 - y/x)/(1 - 4xy^2)
= (2xy^3 - y)/(x - 4x^2 y^2)
again , check my steps
0 answers