What is the derivative of log_{18}(x^2-5x) ?

The derivative of log_{18}(x^2-5x) is (2x-5)/(ln(18)(x^2-5x))

What is the first derivative of log_{18}(x^2-5x) ?

The first derivative of log_{18}(x^2-5x) is (2x-5)/(ln(18)(x^2-5x))

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_{18} (x^{2} - 5 x))

\frac{2 x - 5}{ln ( 18 ) ( x ^{2} - 5 x )}

Solution steps

\frac{d}{d x} (lo g_{18} (x^{2} - 5 x))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( x ^{2} - 5 x )}{ln ( 18 )})

Take the constant out:

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

= \frac{1}{ln ( 18 )} \frac{d}{d x} (ln (x^{2} - 5 x))

Apply the chain rule:

\frac{1}{x ^{2} - 5 x} \frac{d}{d x} (x^{2} - 5 x)

= \frac{1}{x ^{2} - 5 x} \frac{d}{d x} (x^{2} - 5 x)

\frac{d}{d x} (x^{2} - 5 x) = 2 x - 5

= \frac{1}{ln ( 18 )} \cdot \frac{1}{x ^{2} - 5 x} (2 x - 5)

Simplify

\frac{1}{ln ( 18 )} \cdot \frac{1}{x ^{2} - 5 x} (2 x - 5) : \frac{2 x - 5}{ln ( 18 ) ( x ^{2} - 5 x )}

= \frac{2 x - 5}{ln ( 18 ) ( x ^{2} - 5 x )}

\frac{x}{e ^{x} - 1}

\frac{d}{d x} (2 ln (x^{2}))

n = 1 \sum \infty 1 0^{n} x^{n}

\int (x^{2} + 5 x + 6) cos (2 x) d x

y = x, g (y, x) = x^{2}, [0, 1]

What is the derivative of log_{18}(x^2-5x) ?
The derivative of log_{18}(x^2-5x) is (2x-5)/(ln(18)(x^2-5x))
What is the first derivative of log_{18}(x^2-5x) ?
The first derivative of log_{18}(x^2-5x) is (2x-5)/(ln(18)(x^2-5x))