Question

How do you determine dy/dx given `sqrtx+sqrty=25`?

Answer 1

`dy/dx=-sqrt(y/x)`.

Explanation:

`sqrtx+sqrty=25`.

`:. d/dx(sqrtx+sqrty)=d/dx25=0`.

`:. d/dx(sqrtx)+d/dx(sqrty)=0`.

`:. 1/(2sqrtx)+d/dy(sqrty)*dy/dx=0............".[Chain Rule]"`.

`:. 1/(2sqrtx)+1/(2sqrty)*dy/dx=0`

`:. dy/dx=-1/(2sqrtx)/(1/(2sqrty))`

`:. dy/dx=-sqrt(y/x)`.

Answer 2

I found: `(dy)/(dx)=-sqrt(y/x)`

Explanation:

Here you need to use implicit differentiation because your `y` is difficult to be left alone on one side as in our usual functons; but it is not a problem, only remember that `y` is itself a function of `x` and when you differentiate it you need to include the term `(dy)/(dx)` to take into account this dependence; the rest is as usual.

So let us differentiate:

`1/(2sqrt(x))+1/(2sqrt(y))(dy)/(dx)=0`

you see the appearance of the term `(dy)/(dx)` after differentiating `sqrt(y)`!!!

rearrange:

`(dy)/(dx)=-(2sqrt(y))/(2sqrt(x))=-(sqrt(y))/(sqrt(x))=-sqrt(y/x)`