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sandy524Best ResponseYou've already chosen the best response.0
\[f(x)=\sqrt{x^2+2}\]
 one year ago

swissgirlBest ResponseYou've already chosen the best response.1
\( f(x)=(x^2+2)^{1 \over 2}\)
 one year ago

sandy524Best ResponseYou've already chosen the best response.0
thats not one of the choices
 one year ago

swissgirlBest ResponseYou've already chosen the best response.1
noooooooo its easier to solve in this format
 one year ago

rajdoshiBest ResponseYou've already chosen the best response.0
\[1\div2\sqrt{x ^{2}+2x}\times(2x+2)\]
 one year ago

swissgirlBest ResponseYou've already chosen the best response.1
\( f(x)=(x^2+2)^{1 \over 2}\) is the same as \( f(x)= \sqrt{x^2+2}\)
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
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
 one year ago

sandy524Best ResponseYou've already chosen the best response.0
i understand but the assignment is keeping it that way so im getting confused thats all
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
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 ^^
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
swiss is typing up a storm :P she prolly got something for you lol
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
she might just be trolling us, it's hard to tell at this point _ lol
 one year ago

swissgirlBest ResponseYou've already chosen the best response.1
\(f(x)=(x^2+2)^{1 \over 2}\) \( {1 \over 2}(x^2+2)^{({1 \over 2}1)}*2x\) \( {1 \over 2}(x^2+2)^{1 \over 2}*2x\) \(\large { 1 \over 2\sqrt{x^2+2}}*2x\) \(\large {2x \over 2(x^2+2)}\) \(\large {x \over (x^2+2)}\)
 one year ago

swissgirlBest ResponseYou've already chosen the best response.1
okkk i had to post loil but if you are still interested I can explain what i did
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Woops! Your sqrt disappeared on the last couple steps there! <:o
 one year ago

swissgirlBest ResponseYou've already chosen the best response.1
@zepdrix was getting too curious
 one year ago

swissgirlBest ResponseYou've already chosen the best response.1
i hate latex it never listens
 one year ago

swissgirlBest ResponseYou've already chosen the best response.1
ok one more time at it
 one year ago

zepdrixBest ResponseYou've already chosen the best response.1
Do the first 3 steps make sense though sandy? c: those she did correctly.
 one year ago

swissgirlBest ResponseYou've already chosen the best response.1
ok lets explain the first 3 steps that last 2 steps are basic simplification neways
 one year ago

sandy524Best ResponseYou've already chosen the best response.0
u used the power rule first
 one year ago

swissgirlBest ResponseYou've already chosen the best response.1
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
 one year ago

swissgirlBest ResponseYou've already chosen the best response.1
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
 one year ago

swissgirlBest ResponseYou've already chosen the best response.1
Do you follow @sandy524
 one year ago

swissgirlBest ResponseYou've already chosen the best response.1
alrighty ill just give you the last two lines that i messed up \( \large {2x \over 2\sqrt{x^2+2}}\) \( \large {x \over \sqrt{x^2+2}}\)
 one year ago

sandy524Best ResponseYou've already chosen the best response.0
i dont get what this means
 one year ago
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