Question

find the derivative of the function h(x)=sqrt(2-x)

Answer 1

use chain rule

Answer 2

by the sqrt( ) can be written as ( )^(1/2)

Answer 3

we are just in the first section and haven't learned the chain rule, and are required to use a longer method

Answer 4

ok the definition then

Answer 5

right

Answer 6

\[\lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} =f'(x)\]

Answer 7

where we have f(x)=sqrt(2-x)
and so f(x+h)=sqrt(2-(x+h)) <--here just replaced my x's with (x+h)'s

Answer 8

now plug in

Answer 9

sometimes a good method when you are trying to evaluate a limit algebraically and you have a difference/sum of radicals rationalizing might help

Answer 10

to rationalize recall conjugates

Answer 11

for example the conjugate of a-b is a+b
and the conjugate of a+b is a-b

Answer 12

do you want to try to see what happens if you do the following:
\[\lim_{h \rightarrow 0} \frac{\sqrt{2-(x+h)}-\sqrt{2-x}}{h} \cdot \frac{\sqrt{2-(x+h)}+\sqrt{2-x}}{\sqrt{2-(x+h)}+\sqrt{2-x}}\]

Answer 13

right, I got it down to -1/sqrt2-x-h +sqrt2-x

Answer 14

one sec let me check

Answer 15

that looks pretty sweet

Answer 16

\[\lim_{h \rightarrow 0}\frac{-1}{\sqrt{2-x-h}+\sqrt{2-x}}\]

Answer 17

your job is to just replace h with 0

Answer 18

any you could do a little magic afterwards to make the answer look a bit more tidy

Answer 19

so -1/2sqrt2-x?

Answer 20

thank you!

Answer 21

\[\frac{-1}{2 \sqrt{2-x}} \text{ is \right!} \]