## anonymous one year ago find the derivative of the function h(x)=sqrt(2-x)

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1. myininaya

use chain rule

2. myininaya

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

3. anonymous

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

4. myininaya

ok the definition then

5. anonymous

right

6. myininaya

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

7. myininaya

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

8. myininaya

now plug in

9. myininaya

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

10. myininaya

to rationalize recall conjugates

11. myininaya

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

12. myininaya

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

13. anonymous

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

14. myininaya

one sec let me check

15. myininaya

that looks pretty sweet

16. myininaya

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

17. myininaya

your job is to just replace h with 0

18. myininaya

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

19. anonymous

so -1/2sqrt2-x?

20. anonymous

thank you!

21. myininaya

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