# What is dy/dx if square root x = 1+x^2*y?

What is dy/dx if square root x = 1+x^2*y ?

The method required to answer this question is called implicit differentiation:

`d/dx (x)^0.5 =d/dx (1+x^2*y)`  (notice I converted the square root of x to a power)

We begin to differentiate the left hand side:

`(1/2)*(x)^(0.5-1) = d/dx (1+x^2*y)`

And is simplified as follows:

`(1/(2*(x)^0.5)) = d/dx (1)+x^2*y)`

Now that the left hand side is simplified, we need to tackle the right hand side:

`(1/(2*(x)^0.5)) = d/dx (1)+d/dx(x^2*y)`

We know that differentiating a constant will equal zero, and the second term requires the use of the product rule:

`(1/(2*(x)^0.5)) = 0+x^2* dy/dx + 2xy`

Now make dy/dx as the subject of the formula. This is done by taking the terms with dy/dy to the one and the other terms on the other side:

`(1/(2*(x)^0.5)) -2xy = x^2* dy/dx `

`(1/x^2)*((1/(2*(x)^0.5)) -2xy )= dy/dx `

`((1/(2*(x)^(5/2)) -(2y/x) )= dy/dx `

Approved by eNotes Editorial Team