What is the derivative of log_{5}(sqrt(4x^2+5x-8)) ?

The derivative of log_{5}(sqrt(4x^2+5x-8)) is (8x+5)/(2ln(5)(4x^2+5x-8))

What is the first derivative of log_{5}(sqrt(4x^2+5x-8)) ?

The first derivative of log_{5}(sqrt(4x^2+5x-8)) is (8x+5)/(2ln(5)(4x^2+5x-8))

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_{5} (4 x^{2} + 5 x - 8))

\frac{8 x + 5}{2 ln ( 5 ) ( 4 x ^{2} + 5 x - 8 )}

Solution steps

\frac{d}{d x} (lo g_{5} (4 x^{2} + 5 x - 8))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( 4 x ^{2} + 5 x - 8 )}{ln ( 5 )})

Take the constant out:

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

= \frac{1}{ln ( 5 )} \frac{d}{d x} (ln (4 x^{2} + 5 x - 8))

Apply the chain rule:

\frac{1}{4 x ^{2} + 5 x - 8} \frac{d}{d x} (4 x^{2} + 5 x - 8)

= \frac{1}{4 x ^{2} + 5 x - 8} \frac{d}{d x} (4 x^{2} + 5 x - 8)

\frac{d}{d x} (4 x^{2} + 5 x - 8) = \frac{8 x + 5}{2 4 x ^{2} + 5 x - 8}

= \frac{1}{ln ( 5 )} \cdot \frac{1}{4 x ^{2} + 5 x - 8} \cdot \frac{8 x + 5}{2 4 x ^{2} + 5 x - 8}

Simplify

\frac{1}{ln ( 5 )} \cdot \frac{1}{4 x ^{2} + 5 x - 8} \cdot \frac{8 x + 5}{2 4 x ^{2} + 5 x - 8} : \frac{8 x + 5}{2 ln ( 5 ) ( 4 x ^{2} + 5 x - 8 )}

= \frac{8 x + 5}{2 ln ( 5 ) ( 4 x ^{2} + 5 x - 8 )}

\frac{\partial}{\partial x} (x y + \frac{y}{x})

\int \frac{cos ^{5} ( 5 x + 1 )}{sin ^{4} ( 5 x + 1 )} d x

y = \frac{( x - 1 )}{x + 1}

- \frac{5}{7 x + 5}

x \to 0 + lim ((1 + \frac{1}{x})^{x})

What is the derivative of log_{5}(sqrt(4x^2+5x-8)) ?
The derivative of log_{5}(sqrt(4x^2+5x-8)) is (8x+5)/(2ln(5)(4x^2+5x-8))
What is the first derivative of log_{5}(sqrt(4x^2+5x-8)) ?
The first derivative of log_{5}(sqrt(4x^2+5x-8)) is (8x+5)/(2ln(5)(4x^2+5x-8))