babz83
find dx/dy for the following equations
1.) y = square root of 3 (x+1)/x-1)
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babz83
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\[\sqrt[3]{X+1/x-1}\]
babz83
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2) y = cos(lnX)
kushashwa23
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ok
kushashwa23
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let me have a look to this closely
babz83
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please tke your time
Owlfred
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babz83 is smilbourn your mother?
amistre64
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cbrt((x+1)/(x-1)) right?
amistre64
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change the radical to its proper exponent form
[ (x+1)/(x-1) ]^(1/3) and work thru
amistre64
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this aint gonna be nice with my formatting off
amistre64
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lets start with the ^1/3; power rule says bring it down and -1
babz83
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@owlfred...nope remember she said she got an A in calc.... i would not be posting these questions
amistre64
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(1/3) [ (x+1)/(x-1) ]^(-2/3) and out pops the innards for the chain rule
amistre64
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Dx(x+1)/(x-1) = (x-1)(1) - (x+1)(-1) // (x-1)^2 by quotient rule
amistre64
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simplify to: 2x/(x-1)^2 right?
amistre64
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put it all together to go: 2x/3(x-1)^2) [ (x-1)/(x+1) ]^(2/3)
babz83
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am following u amistre
babz83
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let me just digest it!
amistre64
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\[\frac{2x}{3(x-1)^2}\left(\frac{x-1}{x+1}\right)^{2/3}\]right?
amistre64
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ill type it in the latex stuff fo try to make it easier to follow
babz83
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nope
babz83
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\[\sqrt[3]{x+1/x-1}\]
babz83
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ah i get it thats after removing the cube root
amistre64
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\[\sqrt[3]{\frac{x+1}{x-1}}\]
\[Dx\left(\frac{x+1}{x-1}\right)^{1/3}\]
\[\frac{1}{3}\left(\frac{x+1}{x-1}\right)^{-2/3}Dx(\frac{x+1}{x-1})\]
so far
amistre64
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-2/3 just flips it all over and goes positive again right?
babz83
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right
amistre64
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\[\frac{1}{3}\left(\frac{x-1}{x+1}\right)^{2/3}Dx(\frac{x+1}{x-1})\]
\[... Dx(\frac{x+1}{x-1})=\frac{bt'-b't}{b^2}\]
\[... \frac{(x-1)(1)-(x+1)(-1)}{(x-1)^2}\]
\[... \frac{(x-1)+(x+1)}{(x-1)^2}\]
\[... \frac{x-1+x+1}{(x-1)^2}\]
\[... \frac{2x}{(x-1)^2}\]
amistre64
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\[\frac{1}{3}\left(\frac{x-1}{x+1}\right)^{2/3}\frac{2x}{(x-1)^2}\]
are all the parts, just simplify as wanted
amistre64
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i see a spot i missed; (x-1) has a derivative if 1, not -1
babz83
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thnks a lot
babz83
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i see i have a lot of serious reading to do
amistre64
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\[... \frac{(x-1)(1)-(x+1)(1)}{(x-1)^2}\]
\[... \frac{x-1-x-1}{(x-1)^2}\]
\[... \frac{-2}{(x-1)^2}\]
\[\frac{1}{3}\left(\frac{x-1}{x+1}\right)^{2/3}\frac{-2}{(x-1)^2}\]
thats proper, or should be better at least :)