What is the derivative of Log(x) and ln(x)?
Derivative of log (x) depends on the log. The Derivative of THE natural log ln(x) which is log(base)e or e^x is: dy/dx of e^x = e^x dy/dx of e^f(x) = f '(x)e^f(x) dy/dx of lnx = 1/x dy/dx of [ln[f(x)] = f '(x)/f(x)
Derivative of log(3)x ?
Hopefully you worked it andnot merely "found" it. First use change of base formula: y = log_3 x = ln x/ln 3. Since ln 3 is a constant and derivative of ln x is 1/x, dy/dx = 1/[(ln 3)x] as you stated.
What is the derivative of -x * log base 2 (x) - y * log base 2 (y)?
Use implicit differentiation to and the quotient rule: -log base 2 (x)dx - (1/ln2)dx - log base 2 (y)dy - (1/ln2)dy To find the dervative you must set equal to dy/dx to get: (all negatives cancel) dy/dx = (logbase2(x) - (1/ln2))/(logbase2(y) - (1/ln2)) I'm not sure if you can simplify more (or care to).
Find the derivative of y= x log x?
Use the product rule h(x) = f(x)g(x), then h'(x) = f'(x)g(x) + f(x)g'(x) dy/dx = (1/x)(x) + (log(x))(1) = 1 + log(x) assuming (log = natural log), if log is base 10 then dy/dx is 1/ln(10) + log(base10)(x)
What is the derivative of log base 2 (x)?
d/dx (log base a (x)) = 1/ (x ln (a)). So, d/dx (log base 2 (x) = 1/(x ln (2)) As for part 2, the derivative of 2^x d/dx (a^x) = ln(a) * a^x. So, d/dx (2^x) = ln(2) * 2^x