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

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

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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

derivative of

lo g_{4} (x^{2})

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

Solution steps

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

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to \infty lim (7 + \frac{2}{x})

(- 6, 4), (- 11, - 1)

y^{^{'}} = 2 y (1 - y)

\frac{\partial}{\partial x} (\frac{( 2 x + 2 y )}{- 6 x - 8 y})

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