What is the derivative of log_{2}(x-1) ?

The derivative of log_{2}(x-1) is 1/(ln(2)(x-1))

What is the first derivative of log_{2}(x-1) ?

The first derivative of log_{2}(x-1) is 1/(ln(2)(x-1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (lo g_{2} (x - 1))

\frac{1}{ln ( 2 ) ( x - 1 )}

Solution steps

\frac{d}{d x} (lo g_{2} (x - 1))

Apply log rule:

lo g_{a} (b) = \frac{ln ( b )}{ln ( a )}

= \frac{d}{d x} (\frac{ln ( x - 1 )}{ln ( 2 )})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{ln ( 2 )} \frac{d}{d x} (ln (x - 1))

Apply the chain rule:

\frac{1}{x - 1} \frac{d}{d x} (x - 1)

= \frac{1}{x - 1} \frac{d}{d x} (x - 1)

\frac{d}{d x} (x - 1) = 1

= \frac{1}{ln ( 2 )} \cdot \frac{1}{x - 1} \cdot 1

Simplify

\frac{1}{ln ( 2 )} \cdot \frac{1}{x - 1} \cdot 1 : \frac{1}{ln ( 2 ) ( x - 1 )}

= \frac{1}{ln ( 2 ) ( x - 1 )}

x \to - 4 + lim (\frac{2}{x ^{2} - 16})

\int sin^{2} (x) cos^{3} (d) x d x

ln^{2} (8 x^{2})

\int \frac{1}{2} ln x^{2} + 1 d x

\int \frac{sin ( x )}{cos ^{2} ( x ) + 1} d x

What is the derivative of log_{2}(x-1) ?
The derivative of log_{2}(x-1) is 1/(ln(2)(x-1))
What is the first derivative of log_{2}(x-1) ?
The first derivative of log_{2}(x-1) is 1/(ln(2)(x-1))