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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 ?
 one year ago

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

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

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

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

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

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

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

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

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

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

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

UnkleRhaukus Group TitleBest ResponseYou've already chosen the best response.1
right
 one year ago

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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