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Use the product rule to find the derivative of the following y=(3x^2+2)(2x-1)
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If y = f(x).g(x) then dy/dx = f(x).g'(x) + f '(x).g(x) So for this question, dy/dx = (3x^2+2).2 + (6x)(2x-1)
so will it be (9x+2)x(2-1)+(9x+2)(2-1)
Answer is dy/dx = (3x^2+2).2 + (6x)(2x-1)
how did you got that?
Assume f(x)=3x^2+2, so f '(x)=6x + 0 and g(x)=2x-1, so g'(x)= 2 - 0
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Use the product rule for final answer
how did you got 6x+0
derivative of X^n is n.X^(n-1) and derivative of a constant is zero
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