How do you find ##(dy)/(dx)## given ##x^2+y^2=1##?

## dy/dx=-x/y ##

##x^2+y^2=1##

Differentiate wrt ##x##: ## d/dxx^2+d/dxy^2=d/dx1 ##

We already know how to deal with the first and third terms, so lets get them out the way: ## d/dxx^2+d/dxy^2=d/dx1 ## ## :.2x+d/dxy^2=0 ##

For the remaining term we use the , we don’t know how to differentiate ##y^2## wrt ##x## but we do know how to differentiate ##y^2## wrt ##y## (it the same as differentiating ##x^2## wrt ##x##!). The chain rule tells us that: ##dy/dx=dy/(du)*(du)/dx##, so we can rewrite:

## 2x+d/dxy^2=0 ## as ## 2x+d/dy(y^2)*dy/dx=0 ##

We can know perform that final differentiation, as we are now differentiating a function of ##y## wrt ##y## so ## 2x+d/dy(y^2)*dy/dx=0 ## ## :. 2x+2y*dy/dx=0 ## We can then rearrange to get ##dy/dx## as follows: ## 2x+2y*dy/dx=0 ## ## :. 2y*dy/dx=-2x ##

## :. dy/dx=-(2x)/(2y) ##

## :. dy/dx=-x/y ##