Question

How do you implicitly differentiate `x^4-3/y^2-y=8`?

Answer 1

`-(4x^3)/(6y^(-3)-1)`

Explanation:

Here,
`x^4-3/(y^2)-y=8`

now lets differentiate the two sides for x

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

`or,4x^3+6y^(-3)(dy)/(dx)-(dy)/(dx)=0`

`or,(6y^(-3)-1)(dy)/(dx)=-4x^3`

`or,(dy)/(dx)=-(4x^3)/(6y^(-3)-1)`