Differentiate the function y = (x^2)/(3x-1) with respect to x.

This requires use of the quotient rule: d/dx[f(x)/g(x)] = [g(x)f'(x) - g'(x)f(x)]/[g(x)^2]dy/dx = ([(3x-1)*2x] - 3x^2)/[(3x-1)^2],= (3x^2-2x)/[(3x-1)^2],=[x(3x-2)]/[(3x-1)^2]

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Answered by Ted S. Maths tutor

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