Question

How do you implicitly differentiate `3y + y^4/x^2 = 2`?

Thanks for your help!

Answer 1

`(dy)/(dx)=(2xy^4)/(3x^4+4y^3)`

Explanation:

When we implicitly differentiate a function `f(x,y)=0`, whenever we differentiate w.r.t. we use chain rule and when we differentiate w.r.t. `y` and then multiply it by `(dy)/(dx)`.

As such as we have `3y+y^4/x^2=2`

`3xx1xx(dy)/(dx)+(4y^3xx(dy)/(dx)-y^4xx2x)/x^4=0` or

`3(dy)/(dx)+(4y^3(dy)/(dx)-2xy^4)/x^4=0` or

`3(dy)/(dx)+4y^3/x^4(dy)/(dx)-2y^4/x^3=0` or

`(dy)/(dx)[3+(4y^3)/x^4]=(2y^4)/x^3` or

`(dy)/(dx)=((2y^4)/x^3)/(3+(4y^3)/x^4)` or

`(dy)/(dx)=(2xy^4)/(3x^4+4y^3)`