# Math Help - Calculus 4: differentials

1. ## Calculus 4: differentials

Hello everyone,

Could you please tell me if this looks right?

Find the differential of f(x,y)=square root(x^2+y^3) (1,2)

and estimate f(1.04, 1.98)

fx=1/2(x^2+y^3)^-1/2(2x)
fx(1)=.71

fy=1/2(x^2+y^3)^-1/2(3y^s)
fy(2)=1.7

df=fx(a,b)dx+fy(a,b)dy

df=.71dx+1.7dy

f(1.04, 1.98)=square root(1.04^2+1.98^3)=2.

Thank you very much

2. So, we have:

$f(x,y) = \sqrt{x^2 + y^3}$

And we want to estimate:

$f(1.04, 1.98)$

So, we have:

$f_x(x,y) = \frac{1}{2}(x^2 + y^3)^{-\frac{1}{2}}2x$

$=x(x^2 + y^3)^{-\frac{1}{2}}$

And:

$f_y(x,y) = \frac{1}{2}(x^2 + y^3)^{-\frac{1}{2}}3y^2$

$=\frac{3y^2}{2(x^2 + y^3)^{-\frac{1}{2}}}$

So the differential becomes:

$df = x(x^2 + y^3)^{-\frac{1}{2}}dx + \frac{3y^2}{2(x^2 + y^3)^{-\frac{1}{2}}}dy$

Now, how to estimate it... I can't quite remember. I'll get back to you on that if someone else doesn't get to it before me.

3. You just estimate the change in the function using the formula you posted, using x = 1, y = 2, dx = .04 and dy = -.02.