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find dy/dx of the equation y=ln(x)2x^2

Here it is necessary to use the chain rule to solve the derivative. If we equate our equation in terms of the following notation: ln(x)='u'and 2x^2='v' and use the chain rule formula dy/dx=udv/dx+vdu/dx we can solve the derivative:

= lnx(4x)+(2x^2)(1/x) = 4xlnx+2x

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find dy/dx of the equation y=ln(x)2x^2

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