What are the first and second derivatives of #f(x)=ln((4x-5)^(1/3)/(3x+8)^2) #?

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Answer 1 · 2015-11-14T17:52:47

#dy/dx = 4/(3(4x-5)) - 6/(3x + 8)#
#(d^2y)/dx^2 = -8/(3(4x-5)^2) + 9/(3x+8)^2#

Use the log properties to simplify

#y = ln((4x - 5)^(1/3)/(3x+8)^2)#

#y = ln((4x-5)^(1/3))-ln((3x+8)^2)#

#y = ln(4x-5)/3 - 2ln(3x+8)#

Say that #4x - 5 = u# and that #3x + 8 = v#

#y = ln(u)/3 - 2ln(v)#

#dy/dx = 1/3d/(du)ln(u)(du)/dx - 2d/(dv)ln(v)(dv)/dx#

#dy/dx = 1/3*1/ud/dx(4x-5) - 2*1/v*d/dx(3x+8)#

#dy/dx = 1/3*1/u*4 - 2*1/v*3#

#dy/dx = 4/(3(4x-5)) - 6/(3x + 8)#

To find the second derivative, just derivate again, we have

#dy/dx = 4/(3u) - 6/v#

#(d^2y)/dx^2 = 4/3*d/(du)1/u(du)/dx - 6d/(dv)1/v(dv)/dx#

#(d^2y)/dx^2 = 4/3*(-1/(2u^2))*4 - 6\*(-1/(2v^2))*3#

#(d^2y)/dx^2 = -8/(3(4x-5)^2) + 9/(3x+8)^2#