Saturday, September 21, 2019

Calculus: Early Transcendentals, Chapter 3, 3.1, Section 3.1, Problem 22

Differentiate the function y=x(x1)


We have y=x12(x1)y=x12+1x12=x32x12


So,

dydx=ddx(x32x12)=ddx(x32)ddx(x12)=32x32112x121=32x1212x12or=32x12x

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