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farukk Group TitleBest ResponseYou've already chosen the best response.0
which is correct? i.e a+b\ab (after rationallization)a+b\ab * a+b\a+b or a+b\ab * ab\ab ?
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
dw:1354620267974:dw
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
dw:1354620294927:dw
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
can it be ab rather than a+b?
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
it would be equal but it wouldn't helpful in rationalisation
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
all right thanks
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
dw:1354620486122:dw
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
one more question ?
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
dw:1354681791055:dw
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
how will we use power rule in this question?
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
so the first step would be to remember that the the derivative of a sum , is the sum of derivatives
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{\text d}{\text d x}(1+\sqrt{1x^2})=\frac{\text d}{\text d x}(1)+\frac{\text d}{\text d x}\left(\sqrt{1x^2}\right)\]
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
derivative of a constant will be 0
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
so the problem is much simpler now
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
can you remember the derivative of a function of a function?
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
to me it's start right now:)
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
dw:1354682465787:dw
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[ (f\circ g)'(x)=f(g(x))'=f'(g(x))g'(x)\]
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
dw:1354621505691:dw
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
you have to multiply by the derivative of the the inside function
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
dw:1354621612577:dw
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
following?
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
dw:1354682900778:dw
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
\[\frac{\text d}{\text d x}(1+\sqrt{1x^2})\]\[=\frac{\text d}{\text d x}(1)+\frac{\text d}{\text d x}\left(\sqrt{1x^2}\right)\]\[=\frac{\text d}{\text d x}\left(\sqrt{1x^2}\right)\] \[\qquad\frac{\text d}{\text dx}f(g(x))=(f\circ g)'(x)=g'(x)f'(g(x))\]\[\qquad f(x)= x^{1/2}\qquad f'(x)=\frac12x^{1/2}\]\[\qquad g(x)=1x^2\qquad g'(x)=2x\] \[=2x\times\frac12(1x^2)^{1/2}\]
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
i know this stuff is tricky, can you tell me where about you are having trouble
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
it is but u have solved it thank u so much!!
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
yeah but do understand the chain rule/,
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
no will be tomorow in college
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
hey did you use chain rule?
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
the chain rule for the derivative of a function of a function \[(f\circ g)'(x)=g'(x)f'(g(x))\]
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
is there any other way for this question except chain rule?
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
u didt use power rule? did u?
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
i used the power fule to find f '(x) and g'((x)
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
confusing me
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
watch this video , skip the first 5 minutes http://www.khanacademy.org/math/calculus/differentialcalculus/e/chain_rule_1
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
actually i have solved all my questions except this one in quite a simple way
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
did you get the right answers?
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
i'm solving questions using theorems
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
there is no video
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
yeah answer is right
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
well you will have to learn the chain rule
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
this is the link i ment http://www.khanacademy.org/math/calculus/differentialcalculus/v/thechainrule?exid=chain_rule_1
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
dw:1354747831157:dw
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
mistake* 2x\ 2 sqrt 1  x square
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
dw:1354748818882:dw
 2 years ago

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
you can cancel the two
 2 years ago

farukk Group TitleBest ResponseYou've already chosen the best response.0
yeah rightly said!
 2 years ago
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