Answers

2016-02-03T15:43:26+05:30
Y=(x³-3x)²

y=u²                       u=x³-3x
dy/du=2u               du/dx=3x²-3

dy/dx=2u(3x²-3)=6(x³-3x)(x²-1)
1 2 1
Can u give me some basic notes of straight line
equations
please correct your data, u have written 3 in place of 2
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2016-02-03T18:04:03+05:30

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Applying chain rule of differentiation,
d(x³-2x)²/dx =[ d(x³-2x)²/d(x³-2x)]×[d(x³-2x)/dx]
                    =2(x³-2x)×(3x²-2)
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