calyne
Differentiate the function y = ln 2x5x^2



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DHASHNI
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is the ques correct?

calyne
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yup

calyne
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oops

calyne
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fixed it

calyne
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...

amistre64
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\[[ln(u)]'=\frac{u'}{u}\]

calyne
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what are those brackets or 

calyne
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because it's not lnu it's lnu

amistre64
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those are just for grouping; it means that the whole of it is derived and not just the (u) part

calyne
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okay so show me how

amistre64
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i believe it goes piece wise in the end

amistre64
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or you can use the sqrt(u)^2 trick

calyne
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please show me

DHASHNI
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dw:1334421083017:dw

amistre64
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\[\frac{d}{dx}ln(\sqrt{u\ }^2)=\frac{(\sqrt{u\ }^2)'}{ \sqrt{u\ }^2}\to\ \frac{2\sqrt{u\ }*\sqrt{u}'}{ \sqrt{u\ }^2}\]

amistre64
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if i follow that derivatives i think it would look like this ... but best to check with the wolf


amistre64
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\[\frac{2\sqrt{u\ }}{ \sqrt{u\ }^2}\sqrt{u}'\]
\[\frac{2\sqrt{u\ }}{ \sqrt{u\ }^2}\frac{u'}{2\sqrt{u}}=\frac{u'}{ \sqrt{u\ }^2}\to\ \frac{u'}{ u}\]

calyne
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whatevr it's 1/(2x5x^2) * (10x+1) = 10x+1 / 5x^2+x2 woohoo