Calculus

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Differentiate:
y=ln(x/(2x+3))^(1/2)
y=(1/2)ln(x/(2x+3))
y'=0+(1/2)((2x+3-2x)/(2x+3)²)

What's next after the quotient rule?

  • Calculus -

    You don't have to use the quotient rule of differentiation.

    y = lnx - (1/2)ln(2x+3)
    y' = 1/x -(1/2)[1/(2x+3)]*2
    = (1/x) - [1/(2x+3)]
    which can be rewitten with a common denominator, if you wish.

  • Calculus -

    Thanks for your help!

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