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

• Feb 1st 2009, 06:02 PM
Frostking
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)
• Feb 3rd 2009, 09:08 AM
Moo
Hello,
Quote:

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 :
$\displaystyle df=\frac{\partial f}{\partial x} ~dx+\frac{\partial f}{\partial y} ~dy$ (maybe you learnt it another way...)

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

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