Study guides

☆☆

More answers

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

Q: What is the derivative of ln 1 divided by 1-x?

Write your answer...

Submit

Still have questions?

Continue Learning about Other Math

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.

Related questions

1/xlnx Use the chain rule: ln(ln(x)) The derivative of the outside is1/ln(x) times the derivative of the inside. 1/[x*ln(x)]

The first derivative of ln x is 1/x, which (for the following) you better write as x-1.Now use the power rule:Second derivative (the derivative of the first derivative) is -1x-2, the third derivative is the derivative of this, or 2x-3. You may now wish to write this in the alternative form, as 2 / x3.

the derivative of ln x = x'/x; the derivative of 1 is 0 so the answer is 500(1/x)+0 = 500/x

1

It is equal to 0

The derivative of a log is as follows: 1 divided by xlnb Where x is the number beside the log Where b is the base of the log and ln is just the natural log.

y = e^ln x using the fact that e to the ln x is just x, and the derivative of x is 1: y = x y' = 1

The derivative of ln x is 1/x The derivative of 2ln x is 2(1/x) = 2/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).

-1/ln(1-x) * 1/(1-X) or -1/((1-x)*ln(1-x))

The derivative of logx, assuming base 10, is 1/(xln10).

The derivative of ln(x) is 1/x. Therefore, by Chain Rule, we get:[ln(10x)]' = 1/10x * 10 = 1/xUsing this method, you can also infer that the derivative of ln(Ax) where A is any constant equals 1/x.

People also asked