find the derivative of f(x)=sqrt (x^2+2x)

- anonymous

find the derivative of f(x)=sqrt (x^2+2x)

- schrodinger

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- anonymous

\[f(x)=\sqrt{x^2+2}\]

- swissgirl

\( f(x)=(x^2+2)^{1 \over 2}\)

- anonymous

thats not one of the choices

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## More answers

- swissgirl

noooooooo its easier to solve in this format

- anonymous

huh

- anonymous

#4

##### 1 Attachment

- anonymous

\[1\div2\sqrt{x ^{2}+2x}\times(2x+2)\]

- swissgirl

\( f(x)=(x^2+2)^{1 \over 2}\) is the same as \( f(x)= \sqrt{x^2+2}\)

- zepdrix

Sandy, it's easier to differentiate a root if you rewrite it as a fractional exponent :)
She was just rewriting the problem a little nicer for you, not the solution. :D

- anonymous

i understand but the assignment is keeping it that way so im getting confused thats all

- zepdrix

it might be a good idea to switch to a fractional exponent, do the differentiation, then as your last step, put whatever fractional exponent you have back into root form.
UNLESS you remember the derivative of sqrt(x), then you skip doing that step ^^

- zepdrix

swiss is typing up a storm :P she prolly got something for you lol

- anonymous

lol

- zepdrix

she might just be trolling us, it's hard to tell at this point -_- lol

- swissgirl

\(f(x)=(x^2+2)^{1 \over 2}\)
\( {1 \over 2}(x^2+2)^{({1 \over 2}-1)}*2x\)
\( {1 \over 2}(x^2+2)^{-1 \over 2}*2x\)
\(\large { 1 \over 2\sqrt{x^2+2}}*2x\)
\(\large {2x \over 2(x^2+2)}\)
\(\large {x \over (x^2+2)}\)

- swissgirl

okkk i had to post loil but if you are still interested I can explain what i did

- zepdrix

Woops! Your sqrt disappeared on the last couple steps there! <:o

- swissgirl

@zepdrix was getting too curious

- swissgirl

uuugghhhhhhhhhh

- zepdrix

XD

- swissgirl

i hate latex it never listens

- swissgirl

ok one more time at it

- anonymous

:(

- anonymous

ok thanks

- anonymous

you guys rock

- zepdrix

Do the first 3 steps make sense though sandy? c: those she did correctly.

- swissgirl

ok lets explain the first 3 steps that last 2 steps are basic simplification neways

- anonymous

u used the power rule first

- swissgirl

I used the chain rule by first bringing down the exponent in front of the bracket and then subtracting one from the exponent as you can see and then I derived what was in the bracket. The derivative of x^2+2 is 2x

- swissgirl

Well as we see there are 2 things that need to be derived. The outer exponent and the inner exponent. So the chain rule states that you first derive the outer part and then you derive what is in the bracket

- swissgirl

Do you follow @sandy524

- swissgirl

alrighty ill just give you the last two lines that i messed up
\( \large {2x \over 2\sqrt{x^2+2}}\)
\( \large {x \over \sqrt{x^2+2}}\)

- anonymous

i dont get what this means

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