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Mathematics 16 Online
OpenStudy (anonymous):

derivate y=1nx/2x-1

OpenStudy (campbell_st):

is it \[y = 1nx/(2x -1) \] or \[y = \ln(x)/(2x -1)\]

OpenStudy (anonymous):

its y=1nx/(2x-1)

OpenStudy (campbell_st):

well its quotient rule u = 1nx du/dx = n v = 2x -1 dv/dx = 2 \[dy/dx =( v \times du/dx - u \times dv/dx)/(v)^2\] \[dy/dx = ((1nx)\times 2 - (2x -1)\times n)/(1nx)^2\]

OpenStudy (campbell_st):

this simplifes \[dy/dx = (2nx - 2nx + n)/n^2x^2 = 1/nx^2\]

OpenStudy (anonymous):

thanx so much!

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