Math (Calculus)

y= x^(2x+1)

find the derivative of the function using natural log

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asked by Amanda
  1. Log(y) = (2x+1)Log(x)

    Differentiate both sides:

    y'/y = 2 Log(x) + 2 + 1/x


    Multiply by y:

    y' = [2 Log(x) + 2 + 1/x] x^(2x+1)

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  2. y=x2x+1


    Find the derivative of the expression.

    (2x+1)


    To find the derivative of 2x, multiply the base (x) by the exponent (1), then subtract 1 from the exponent (1-1=0). Since the exponent is now 0, x is eliminated from the term.

    1


    Since 1 does not contain x, the derivative of 1 is 0.

    2x+1=2+0


    Add 0 to 2 to get 2.

    2x+1=2


    Using the chain rule, the derivative of x2x+1 is

    x2x+1*2


    Multiply x2x+1 by 2 to get 2x2x+1.

    2x2x+1


    The derivative of y with respect to x is 2x2x+1.

    The derivative of y with respect to x is 2x2x+1.

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