Question

How do you find `dy/dx` by implicit differentiation of `x^3+y^3=4xy+1` and evaluate at point (2,1)?

Answer 1

`8/5`

Explanation:

`d/(dx)(x^3+y^3=4xy+1)--(1)`

we will differentiate `wrt" "x`. Remembering that when differentiating `y` we multiply by `(dy)/(dx)` by virtue of the chain rule. Also on the `RHS` we will need the product rule on the first term

`(1)rarr3x^2+3y^2(dy)/(dx)=4y+4x(dy)/(dx)+0`

rearrange for `(dy)/(dx)`

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

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

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

`:.[(dy)/(dx)]_(color(white)(=)(2,1))=(4xx1-3xx2^2)/(3xx1^2-4xx2`

`=(4-12)/(3-8)=-8/-5=8/5`