What is the derivative of 4log_{3}(x^2-1) ?

The derivative of 4log_{3}(x^2-1) is (8x)/(ln(3)(x^2-1))

What is the first derivative of 4log_{3}(x^2-1) ?

The first derivative of 4log_{3}(x^2-1) is (8x)/(ln(3)(x^2-1))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

\frac{d}{d x} (4 lo g_{3} (x^{2} - 1))

\frac{8 x}{ln ( 3 ) ( x ^{2} - 1 )}

Solution steps

\frac{d}{d x} (4 lo g_{3} (x^{2} - 1))

Take the constant out:

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

= 4 \frac{d}{d x} (lo g_{3} (x^{2} - 1))

Apply log rule:

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

= 4 \frac{d}{d x} (\frac{ln ( x ^{2} - 1 )}{ln ( 3 )})

Take the constant out:

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

= 4 \cdot \frac{1}{ln ( 3 )} \frac{d}{d x} (ln (x^{2} - 1))

Apply the chain rule:

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

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

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

= 4 \cdot \frac{1}{ln ( 3 )} \cdot \frac{1}{x ^{2} - 1} \cdot 2 x

Simplify

4 \cdot \frac{1}{ln ( 3 )} \cdot \frac{1}{x ^{2} - 1} \cdot 2 x : \frac{8 x}{ln ( 3 ) ( x ^{2} - 1 )}

= \frac{8 x}{ln ( 3 ) ( x ^{2} - 1 )}

\frac{d}{d x} (20 x)

\int \frac{1}{x ( 1 + 4 I n ^{2} x )} d x

y = 16 x

f (v) = \frac{v}{v + \frac{d}{v}}

y = 3 x, y = \frac{3}{2} x

What is the derivative of 4log_{3}(x^2-1) ?
The derivative of 4log_{3}(x^2-1) is (8x)/(ln(3)(x^2-1))
What is the first derivative of 4log_{3}(x^2-1) ?
The first derivative of 4log_{3}(x^2-1) is (8x)/(ln(3)(x^2-1))