Find the derivative of f(x)=log(7-2x)?

Please read.

f(x) = ln(7 - 2x)

(I'm assuming the natural log here, but any other logarithm base simply differs by a multiplicative constant).

Let u = 7 - 2x, then f = ln(u), and by the chain rule, df/dx = df/du*du/dx

du/dx -2

df/du = 1/u

So:

df/dx = -2/u = -2/(7 - 2x) = 2/(2x - 7)

Assuming that it is a decimal log (base 10), we have:

f'(x) = [log(7 - 2x)]' * (7 - 2x)' = 1/[(7 - 2x) * ln (10)] * (-2) = -2/[(7 - 2x) * ln (10)].


Or, alternatively: f'(x) = 2/[(2x - 7) * ln (10)] (we multiplied both the numerator and the denominator by -1).


P.s.: If the base of the logarithm equals e, the derivative simplifies to:

f'(x) = 2/[(2x - 7) * ln (e)] = 2/[(2x - 7) * 1] = 2/[(2x - 7)].

f'(x) = -2/((7-2x)ln10)

Let y = log u
u = 7 - 2x
du/dx = - 2
dy/du = 1/u

dy/dx= - 2 / u = - 2 / [ 7 - 2x ]

-2/(7-2x). It is the chain rule. the derivative of log(x) is 1/x, so then by the chain rule the derivative of (im going to cal g(x)=7-2x) log(g(x)) is (1/g(x)) times the derivative of g(x). So here 1/gx) is 1/(7-2x), and the derivative of g(x) is -2, so the derivative of f(x)=log(g(x))=-2/(7-2x)