sandy524
find the derivative of f(x)=sqrt (x^2+2x)
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sandy524
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\[f(x)=\sqrt{x^2+2}\]
swissgirl
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\( f(x)=(x^2+2)^{1 \over 2}\)
sandy524
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thats not one of the choices
swissgirl
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noooooooo its easier to solve in this format
sandy524
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huh
sandy524
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#4
rajdoshi
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\[1\div2\sqrt{x ^{2}+2x}\times(2x+2)\]
swissgirl
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\( f(x)=(x^2+2)^{1 \over 2}\) is the same as \( f(x)= \sqrt{x^2+2}\)
zepdrix
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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
sandy524
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i understand but the assignment is keeping it that way so im getting confused thats all
zepdrix
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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
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swiss is typing up a storm :P she prolly got something for you lol
sandy524
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lol
zepdrix
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she might just be trolling us, it's hard to tell at this point -_- lol
swissgirl
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\(f(x)=(x^2+2)^{1 \over 2}\)
\( {1 \over 2}(x^2+2)^{({1 \over 2}-1)}*2x\)
\( {1 \over 2}(x^2+2)^{-1 \over 2}*2x\)
\(\large { 1 \over 2\sqrt{x^2+2}}*2x\)
\(\large {2x \over 2(x^2+2)}\)
\(\large {x \over (x^2+2)}\)
swissgirl
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okkk i had to post loil but if you are still interested I can explain what i did
zepdrix
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Woops! Your sqrt disappeared on the last couple steps there! <:o
swissgirl
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@zepdrix was getting too curious
swissgirl
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uuugghhhhhhhhhh
zepdrix
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XD
swissgirl
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i hate latex it never listens
swissgirl
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ok one more time at it
sandy524
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:(
sandy524
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ok thanks
sandy524
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you guys rock
zepdrix
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Do the first 3 steps make sense though sandy? c: those she did correctly.
swissgirl
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ok lets explain the first 3 steps that last 2 steps are basic simplification neways
sandy524
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u used the power rule first
swissgirl
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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
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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
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Do you follow @sandy524
swissgirl
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alrighty ill just give you the last two lines that i messed up
\( \large {2x \over 2\sqrt{x^2+2}}\)
\( \large {x \over \sqrt{x^2+2}}\)
sandy524
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i dont get what this means