What is the derivative of log_{10}(x^2-5x) ?

The derivative of log_{10}(x^2-5x) is (2x-5)/(ln(10)(x^2-5x))

What is the first derivative of log_{10}(x^2-5x) ?

The first derivative of log_{10}(x^2-5x) is (2x-5)/(ln(10)(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_{10} (x^{2} - 5 x))

\frac{2 x - 5}{ln ( 10 ) ( x ^{2} - 5 x )}

Solution steps

\frac{d}{d x} (lo g_{10} (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 ( 10 )})

Take the constant out:

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

= \frac{1}{ln ( 10 )} \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 ( 10 )} \cdot \frac{1}{x ^{2} - 5 x} (2 x - 5)

Simplify

\frac{1}{ln ( 10 )} \cdot \frac{1}{x ^{2} - 5 x} (2 x - 5) : \frac{2 x - 5}{ln ( 10 ) ( x ^{2} - 5 x )}

= \frac{2 x - 5}{ln ( 10 ) ( x ^{2} - 5 x )}

y = x (x - 6)^{2}

e^{- 5 x} + e^{3 x}

\frac{d}{d x} (e^{x} + 19)

x \to 2 + lim (2 - x)

\frac{d}{d x} (\frac{1 - 6 x}{2 + x})

What is the derivative of log_{10}(x^2-5x) ?
The derivative of log_{10}(x^2-5x) is (2x-5)/(ln(10)(x^2-5x))
What is the first derivative of log_{10}(x^2-5x) ?
The first derivative of log_{10}(x^2-5x) is (2x-5)/(ln(10)(x^2-5x))