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Blakeln(1)/[1-x]?
d/dx(u/v)=(v*du/dx-u*dv/dx)/(v2)
d/dx(ln(1)/[1-x])=[(1-x)*d/dx(ln1)-ln1*d/dx(1-x)]/[(1-x)2]
-The derivative of ln1 is:
d/dx(lnu)=(1/u)*d/dx(u)
d/dx(ln1)=(1/1)*d/dx(1)
d/dx(ln1)=(1)*d/dx(1)
d/dx(ln1)=d/dx(1)
-The derivative of 1-x is:
d/dx(u-v)=du/dx-dv/dx
d/dx(1-x)=d/dx(1)-d/dx(x)
d/dx(ln(1)/[1-x])=[(1-x)*d/dx(1)-ln1*(d/dx(1)-d/dx(x))]/[(1-x)2]
-The derivative of 1 is 0 because it is a constant.
-The derivative of x is:
d/dx(xn)=nxn-1
d/dx(x)=1*x1-1
d/dx(x)=1*x0
d/dx(x)=1*(1)
d/dx(x)=1
d/dx(ln(1)/[1-x])=[(1-x)*(0)-ln1*(0-1)]/[(1-x)2]
d/dx(ln(1)/[1-x])=[-ln1*(-1)]/[(1-x)2]
d/dx(ln(1)/[1-x])=[ln1]/[(1-x)2]
But you see, ln1 is equal to 0:
d/dx(ln(1)/[1-x])=[0]/[(1-x)2]
d/dx(ln(1)/[1-x])=0
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