Question

How do you determine dy/dx given `ysqrtx+xsqrty=16`?

Answer 1

`y'=(-y/(2sqrtx)-sqrty)/(sqrtx+(x)/(2sqrty))`

Explanation:

We have to use implicit differentiation which is basically just a special case of the chain rule. Every time we differentiate a factor that has a `y` variable we have to include the factor `dy/dx` or `y'`.

We will also be using the product rule and power rule to answer this question.

Begin by changing all of the radicals to their exponent equivalents.

`yx^(1/2)+xy^(1/2)=16`

`y(1/2)x^(-1/2)+(1)y'x^(1/2)+x(1/2)y^(-1/2)y'+(1)y^(1/2)=0`

Simplify and switch back to radicals

`y/(2sqrtx)+y'sqrtx+(xy')/(2sqrty)+sqrty=0`

Gather terms with `y'` to one side and everything else to the other side

`y'sqrtx+(xy')/(2sqrty)=-y/(2sqrtx)-sqrty`

Factor out `y'`

`y'(sqrtx+(x)/(2sqrty))=-y/(2sqrtx)-sqrty`

Isolate `y'` by dividing

`y'cancel(sqrtx+(x)/(2sqrty))/cancel(sqrtx+(x)/(2sqrty))=(-y/(2sqrtx)-sqrty)/(sqrtx+(x)/(2sqrty))`

`y'=(-y/(2sqrtx)-sqrty)/(sqrtx+(x)/(2sqrty))`

Click on the tutorial to see another example.