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OCW Scholar - Single Variable Calculus
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Derivative of y = 6(x)ln(x)-x+1? I haven't done any calculus in a year or so but think it's something to do with the chain rule. Not sure what to choose as the "inside" and "outside" parts though - any help very much appreciated.
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If im interpreting your equation right its: \[y=(6x)*ln(x)-x+1, \right?\] if you use the chain rule and make ln(x) the inside function of \[U^V\], where lnx=U and 6x=V, you risk getting something ugly such as ln(ln(x)) in your unsimplified answer, which is ugly. try the product rule with the same assignments, U=lnx and V=x U'=1/x and V'=6 so U'V+UV' = (1/x)6x+6lnx in total we get 6lnx+6-1 or 6lnx+5
Aha, of course it was the product rule I should have been using! Great explanation, thank you :)
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