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anonymous
 4 years ago
find the derivative of f(x)=sqrt (x^2+2x)
anonymous
 4 years ago
find the derivative of f(x)=sqrt (x^2+2x)

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[f(x)=\sqrt{x^2+2}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\( f(x)=(x^2+2)^{1 \over 2}\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0thats not one of the choices

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0noooooooo its easier to solve in this format

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[1\div2\sqrt{x ^{2}+2x}\times(2x+2)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\( f(x)=(x^2+2)^{1 \over 2}\) is the same as \( f(x)= \sqrt{x^2+2}\)

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Sandy, 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
 4 years ago
Best ResponseYou've already chosen the best response.0i understand but the assignment is keeping it that way so im getting confused thats all

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1it 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
 4 years ago
Best ResponseYou've already chosen the best response.1swiss is typing up a storm :P she prolly got something for you lol

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\(f(x)=(x^2+2)^{1 \over 2}\) \( {1 \over 2}(x^2+2)^{({1 \over 2}1)}*2x\) \( {1 \over 2}(x^2+2)^{1 \over 2}*2x\) \(\large { 1 \over 2\sqrt{x^2+2}}*2x\) \(\large {2x \over 2(x^2+2)}\) \(\large {x \over (x^2+2)}\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0okkk i had to post loil but if you are still interested I can explain what i did

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Woops! Your sqrt disappeared on the last couple steps there! <:o

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@zepdrix was getting too curious

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i hate latex it never listens

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok one more time at it

zepdrix
 4 years ago
Best ResponseYou've already chosen the best response.1Do the first 3 steps make sense though sandy? c: those she did correctly.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok lets explain the first 3 steps that last 2 steps are basic simplification neways

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0u used the power rule first

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Do you follow @sandy524

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0alrighty 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
 4 years ago
Best ResponseYou've already chosen the best response.0i dont get what this means
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