# Thread: differential of (x^2 + y^3)^1/2 at (1,2)

1. ## differential of (x^2 + y^3)^1/2 at (1,2)

We just started partial derivatives in vector calculus and now we are being introduced to differentials.

I need to find the differential of (x^2 + y^3)^1/2 at the point (1,2)

I know that differential is equal to the sum of both partial derivatives so is the answer

3y^2 / (2(y^3 + x^2)^(1/2) + x/ (y^3 + x^2)^(1/2)

2. Hello,
Originally Posted by Frostking
We just started partial derivatives in vector calculus and now we are being introduced to differentials.

I need to find the differential of (x^2 + y^3)^1/2 at the point (1,2)

I know that differential is equal to the sum of both partial derivatives so is the answer

3y^2 / (2(y^3 + x^2)^(1/2) + x/ (y^3 + x^2)^(1/2)
Actually, the differential df of a function is :
$df=\frac{\partial f}{\partial x} ~dx+\frac{\partial f}{\partial y} ~dy$ (maybe you learnt it another way...)

So $df_{(x,y)}=\frac{x}{\sqrt{y^3+x^2}} ~dx+\frac{3y^2}{2 \sqrt{y^3+x^2}} ~dy$ (as you correctly calculated)

Now let $x=1$ and $y=2$