anonymous
  • anonymous
???
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
which is correct? i.e a+b\a-b (after rationallization)a+b\a-b * a+b\a+b or a+b\a-b * a-b\a-b ?
UnkleRhaukus
  • UnkleRhaukus
|dw:1354620267974:dw|
UnkleRhaukus
  • UnkleRhaukus
|dw:1354620294927:dw|

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More answers

anonymous
  • anonymous
can it be a-b rather than a+b?
UnkleRhaukus
  • UnkleRhaukus
it would be equal but it wouldn't helpful in rationalisation
anonymous
  • anonymous
all right thanks
UnkleRhaukus
  • UnkleRhaukus
|dw:1354620486122:dw|
anonymous
  • anonymous
one more question ?
anonymous
  • anonymous
may i ask
UnkleRhaukus
  • UnkleRhaukus
yes!
anonymous
  • anonymous
|dw:1354681791055:dw|
anonymous
  • anonymous
u got?
anonymous
  • anonymous
how will we use power rule in this question?
UnkleRhaukus
  • UnkleRhaukus
so the first step would be to remember that the the derivative of a sum , is the sum of derivatives
anonymous
  • anonymous
yes yes
UnkleRhaukus
  • UnkleRhaukus
\[\frac{\text d}{\text d x}(1+\sqrt{1-x^2})=\frac{\text d}{\text d x}(1)+\frac{\text d}{\text d x}\left(\sqrt{1-x^2}\right)\]
anonymous
  • anonymous
derivative of a constant will be 0
UnkleRhaukus
  • UnkleRhaukus
right
UnkleRhaukus
  • UnkleRhaukus
so the problem is much simpler now
UnkleRhaukus
  • UnkleRhaukus
can you remember the derivative of a function of a function?
anonymous
  • anonymous
to me it's start right now:)
anonymous
  • anonymous
i'm sorry
anonymous
  • anonymous
|dw:1354682465787:dw|
UnkleRhaukus
  • UnkleRhaukus
\[ (f\circ g)'(x)=f(g(x))'=f'(g(x))g'(x)\]
UnkleRhaukus
  • UnkleRhaukus
|dw:1354621505691:dw|
UnkleRhaukus
  • UnkleRhaukus
you have to multiply by the derivative of the the inside function
UnkleRhaukus
  • UnkleRhaukus
|dw:1354621612577:dw|
UnkleRhaukus
  • UnkleRhaukus
following?
anonymous
  • anonymous
|dw:1354682900778:dw|
UnkleRhaukus
  • UnkleRhaukus
yeah
anonymous
  • anonymous
but how?
UnkleRhaukus
  • UnkleRhaukus
\[\frac{\text d}{\text d x}(1+\sqrt{1-x^2})\]\[=\frac{\text d}{\text d x}(1)+\frac{\text d}{\text d x}\left(\sqrt{1-x^2}\right)\]\[=\frac{\text d}{\text d x}\left(\sqrt{1-x^2}\right)\] \[\qquad\frac{\text d}{\text dx}f(g(x))=(f\circ g)'(x)=g'(x)f'(g(x))\]\[\qquad f(x)= x^{1/2}\qquad f'(x)=\frac12x^{-1/2}\]\[\qquad g(x)=1-x^2\qquad g'(x)=-2x\] \[=-2x\times\frac12(1-x^2)^{-1/2}\]
UnkleRhaukus
  • UnkleRhaukus
i know this stuff is tricky, can you tell me where about you are having trouble
anonymous
  • anonymous
haha
anonymous
  • anonymous
it is but u have solved it thank u so much!!
UnkleRhaukus
  • UnkleRhaukus
yeah but do understand the chain rule/,
anonymous
  • anonymous
no will be tomorow in college
anonymous
  • anonymous
hey did you use chain rule?
UnkleRhaukus
  • UnkleRhaukus
the chain rule for the derivative of a function of a function \[(f\circ g)'(x)=g'(x)f'(g(x))\]
anonymous
  • anonymous
is there any other way for this question except chain rule?
anonymous
  • anonymous
u didt use power rule? did u?
UnkleRhaukus
  • UnkleRhaukus
i used the power fule to find f '(x) and g'((x)
anonymous
  • anonymous
confusing me
UnkleRhaukus
  • UnkleRhaukus
watch this video , skip the first 5 minutes http://www.khanacademy.org/math/calculus/differential-calculus/e/chain_rule_1
anonymous
  • anonymous
actually i have solved all my questions except this one in quite a simple way
UnkleRhaukus
  • UnkleRhaukus
did you get the right answers?
anonymous
  • anonymous
i'm solving questions using theorems
anonymous
  • anonymous
there is no video
anonymous
  • anonymous
yeah answer is right
UnkleRhaukus
  • UnkleRhaukus
well you will have to learn the chain rule
UnkleRhaukus
  • UnkleRhaukus
this is the link i ment http://www.khanacademy.org/math/calculus/differential-calculus/v/the-chain-rule?exid=chain_rule_1
anonymous
  • anonymous
|dw:1354747831157:dw|
anonymous
  • anonymous
mistake* -2x\ 2 sqrt 1 - x square
anonymous
  • anonymous
|dw:1354748818882:dw|
UnkleRhaukus
  • UnkleRhaukus
you can cancel the two
anonymous
  • anonymous
yeah rightly said!

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