DLS
  • DLS
DE help
Mathematics
katieb
  • katieb
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DLS
  • DLS
\[y=\log(x^{2}\sqrt{x^{2}+1})\]
DLS
  • DLS
I got something like this \[\frac{1}{x^{2}\sqrt{x^{2}+1}} \times 2\sqrt{x^{2}+1} \times 2x\]
DLS
  • DLS

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ghazi
  • ghazi
there is a bit of editing you should have used product rule
ghazi
  • ghazi
for (x^2 sqrt{x^2+1})
DLS
  • DLS
how?:o
ghazi
  • ghazi
\[\frac{ d(x^2 \sqrt{x^2+1}) }{ dx }=x^2\frac{ d \sqrt{x^2+1} }{ dx }+\sqrt{x^2+1}*2x\]
DLS
  • DLS
woah :o thanks
ghazi
  • ghazi
:) YW
DLS
  • DLS
how did you get dx in the denominator?
ghazi
  • ghazi
you are differentiating with respect to x
ghazi
  • ghazi
i can write it as\[\frac{ d }{ dx }(\sqrt{x^2+1}*x^2)\]
DLS
  • DLS
oh wait equation is messed up lol just asec
DLS
  • DLS
\[\frac{dy}{dx}=\frac{dlog(x^{2}\sqrt{x^{2}+1}}{dx^{2}\sqrt{x^{2}+1}} \times \frac{dx^{2} \sqrt{x^{2}+1}}{d \sqrt{x^{2}+1}} \] ..so on
ghazi
  • ghazi
NO
DLS
  • DLS
i do like this,where i didnt diff. wrt. dx
DLS
  • DLS
thats how im taught >.>
ghazi
  • ghazi
you have to differentiate wrt x, you can't diffrentiate wrt to function
DLS
  • DLS
wth :/ am i taught wrong :S but i was getting all the answers like this
ghazi
  • ghazi
differentiation is done wrt to a variable basically differentiation tells us rate of change of one variable wrt to another and function is something that tells us the relation between those two variables so we can never get rate of change wrt to a function
DLS
  • DLS
okay..!
ghazi
  • ghazi
\[\frac{ d }{ dx }(\sqrt{x^2+1})*x^2+2x* \sqrt{x^2+1}\]
DLS
  • DLS
okay...clear!
ghazi
  • ghazi
cool :D
ghazi
  • ghazi
\[\frac{ d }{ dx } (\sqrt{x^2+1})=\frac{ 1 }{ \sqrt{x^2+1} }*\frac{ d }{ dx }(x^2+1)\]
ghazi
  • ghazi
there will be a factor of -1/2 multiplied in RHS
DLS
  • DLS
derivative of sqrt x is 1/2x right :o
ghazi
  • ghazi
-1/2 x
DLS
  • DLS
\[\frac{d \sqrt{x}}{dx}= \frac{1}{2 \sqrt{x}}\] does my coaching class suck lol :/
DLS
  • DLS
http://wiki.answers.com/Q/What_is_the_derivative_of_square_root_of_x
ghazi
  • ghazi
no they are right let me show you
DLS
  • DLS
will it be like this: \[\frac{1}{2 \sqrt{x^{2}+1}} \times 2x = \frac{x}{\sqrt{x^{2}+1}} \]
ghazi
  • ghazi
\[\frac{ d }{ dx } (x)^{-1/2}= \frac{ -1 }{ 2 }(x)^{-1/2-1}\]
DLS
  • DLS
so how do we have 2 answers
ghazi
  • ghazi
no we have one, just be careful in calculating power
DLS
  • DLS
okay..
DLS
  • DLS
so final answer im getting is
DLS
  • DLS
\[\frac{1}{x(x^{2}+1}\]
DLS
  • DLS
\[\frac{1}{x^{2} \sqrt{x^{2}+1}} \times \frac{x}{\sqrt{x^{2}+1}}\]
DLS
  • DLS
which is wrong :/ http://www.wolframalpha.com/input/?i=derivative+of+log%28x%5E2%28sqrt%28x%5E2%2B1%29%29%29
ghazi
  • ghazi
\[\frac{ d }{ dx } (\sqrt{x^2+1})=\frac{1 }{ 2 }(\sqrt{x^2+1})^{-3/2}* 2x\] your answer
DLS
  • DLS
what did you do
ghazi
  • ghazi
\[\frac{ d }{ dx }(x^n)=nx^{n-1}\] hope you know this formula
DLS
  • DLS
oh okay!

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