Approximating (24.5)^(1/2) + (9.5)^(1/2) using Differentials

[Nachore]

1. The problem
Use Differentials to approximate (24.5)^(1/2) + (9.5)^(1/2). Compare your answer to your calculator's answer.

Homework Equations

I used z = (x)^(1/2) + (y)^(1/2)

The Attempt at a Solution

What I used:
let z = (x)^(1/2) + (y)^(1/2)
x = 25
y = 10
dx = 0.5
dy = 0.5
Using http://www.maths.soton.ac.uk/~jav/soton/ma155/lectures2/img203.gif .[/URL]
I plugged in 1/(2(x)^(1/2)) for the partial derivative w/respect to x, and 1/(2(y)^(1/2)) for the partial derivative w/respect to y. When I plug into the formula, I get 0.1290 not 8.0139 like in the calculator. What am I doing wrong?

 

[HallsofIvy]

1. The problem
Use Differentials to approximate (24.5)^(1/2) + (9.5)^(1/2). Compare your answer to your calculator's answer.

Homework Equations

I used z = (x)^(1/2) + (y)^(1/2)

The Attempt at a Solution

What I used:
let z = (x)^(1/2) + (y)^(1/2)
x = 25
y = 10
dx = 0.5
dy = 0.5
Using http://www.maths.soton.ac.uk/~jav/soton/ma155/lectures2/img203.gif .[/URL]
I plugged in 1/(2(x)^(1/2)) for the partial derivative w/respect to x, and 1/(2(y)^(1/2)) for the partial derivative w/respect to y. When I plug into the formula, I get 0.1290 not 8.0139 like in the calculator. What am I doing wrong?

You the did the derivative wrong. The derivative of xⁿ is n x^n-1, not n xⁿ as you have.