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The derivative of ln x, the natural logarithm, is 1/x.

Otherwise, given the identity logbx = log(x)/log(b), we know that the derivative of logbx = 1/(x*log b).

ProofThe derivative of ln x follows quickly once we know that the derivative of ex is itself.Let y = ln x (we're interested in knowing dy/dx)

Then ey = x

Differentiate both sides to get ey dy/dx = 1

Substitute ey = x to get x dy/dx = 1, or dy/dx = 1/x.

Differentiation of log (base 10) x

log (base 10) x

= log (base e) x * log (base 10) e

d/dx [ log (base 10) x ]

= d/dx [ log (base e) x * log (base 10) e ]

= [log(base 10) e] / x

= 1 / x ln(10)

Q: Differentiate log x

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Log (x^3) = 3 log(x) Log of x to the third power is three times log of x.

Here are a few, note x>0 and y>0 and a&b not = 1 * log (xy) = log(x) + log(y) * log(x/y) = log(x) - log(y) * loga(x) = logb(x)*loga(b) * logb(bn) = n * log(xa) = a*log(x) * logb(b) = 1 * logb(1) = 0

log(9x) + log(x) = 4log(10)log(9) + log(x) + log(x) = 4log(10)2log(x) = 4log(10) - log(9)log(x2) = log(104) - log(9)log(x2) = log(104/9)x2 = 104/9x = 102/3x = 33 and 1/3

log base 2 of [x/(x - 23)]

1

Related questions

The answer is ln(2)2x where ln(2) is the natural log of 2. The answer is NOT f(x) = x times 2 to the power(x-1). That rule applies only when the exponent is a constant.

log(x6) = log(x) + log(6) = 0.7782*log(x) log(x6) = 6*log(x)

Log (x^3) = 3 log(x) Log of x to the third power is three times log of x.

log(x) - log(6) = log(15)Add log(6) to each side:log(x) = log(15) + log(6) = log(15 times 6)x = 15 times 6x = 90

Here are a few, note x>0 and y>0 and a&b not = 1 * log (xy) = log(x) + log(y) * log(x/y) = log(x) - log(y) * loga(x) = logb(x)*loga(b) * logb(bn) = n * log(xa) = a*log(x) * logb(b) = 1 * logb(1) = 0

log x + 2 = log 9 log x - log 9 = -2 log (x/9) = -2 x/9 = 10^(-2) x/9 = 1/10^2 x/9 = 1/100 x= 9/100 x=.09

log(x) + log(2) = log(2)Subtract log(2) from each side:log(x) = 0x = 100 = 1

log(9x) + log(x) = 4log(10)log(9) + log(x) + log(x) = 4log(10)2log(x) = 4log(10) - log(9)log(x2) = log(104) - log(9)log(x2) = log(104/9)x2 = 104/9x = 102/3x = 33 and 1/3

2*log(15) = log(x) 152 = x; its equivalent logarithmic form is 2 = log15 x (exponents are logarithms) then, it is equivalent to 2log 15 = log x, equivalent to log 152 = log x (the power rule), ... 2 = log15 x 2 = log x/log 15 (using the change-base property) 2log 15 = log x Thus, we can say that 152 = x is equivalent to 2*log(15) = log(x) (equivalents to equivalents are equivalent)

log base 2 of [x/(x - 23)]

5x 12x = 17xx log(5) + x log(12) = x log(17)x [ log(5) + log(12) ] = x log(17)x log(60) = x log(17)x = 0This actually checks. Since anything to the zero power is ' 1 ',50 120 = 1 times 1, or 1and 170 = 1

log(1) = 0log(x*y) = log(x) + log(y)If logb(x) = y then x = by.