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ln(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
the derivative of ln x = x'/x; the derivative of 1 is 0 so the answer is 500(1/x)+0 = 500/x
The derivative of ln x is 1/x. Replacing the expression, that gives you 1 / (1-x). By the chain rule, this must then be multiplied by the derivative of (1-x), which is -1. So, the final result is -1 / (1-x).
The antiderivative of x2 + x is 1/3x3 + 1/2x2 + C.
Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.Yes, the function ln(x) where ln is the logarithm to base e.
Assuming that is the natural logarithm (logarithm to base e), the derivative of ln x is 1/x. For other bases, the derivative of logax = 1 / (x ln a), where ln a is the natural logarithm of a. Natural logarithms are based on the number e, which is approximately 2.718.