What is the derivative of log_{5}(x^4sqrt(2x^2+9x)) ?

The derivative of log_{5}(x^4sqrt(2x^2+9x)) is (20x+81)/(2ln(5)x(2x+9))

What is the first derivative of log_{5}(x^4sqrt(2x^2+9x)) ?

The first derivative of log_{5}(x^4sqrt(2x^2+9x)) is (20x+81)/(2ln(5)x(2x+9))

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} (x^{4} 2 x^{2} + 9 x))

\frac{20 x + 81}{2 ln ( 5 ) x ( 2 x + 9 )}

Solution steps

\frac{d}{d x} (lo g_{5} (x^{4} 2 x^{2} + 9 x))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( x ^{4} 2 x ^{2} + 9 x )}{ln ( 5 )})

Take the constant out:

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

= \frac{1}{ln ( 5 )} \frac{d}{d x} (ln (x^{4} 2 x^{2} + 9 x))

Apply the chain rule:

\frac{1}{x ^{4} 2 x ^{2} + 9 x} \frac{d}{d x} (x^{4} 2 x^{2} + 9 x)

= \frac{1}{x ^{4} 2 x ^{2} + 9 x} \frac{d}{d x} (x^{4} 2 x^{2} + 9 x)

\frac{d}{d x} (x^{4} 2 x^{2} + 9 x) = \frac{20 x ^{5} + 81 x ^{4}}{2 2 x ^{2} + 9 x}

= \frac{1}{ln ( 5 )} \cdot \frac{1}{x ^{4} 2 x ^{2} + 9 x} \cdot \frac{20 x ^{5} + 81 x ^{4}}{2 2 x ^{2} + 9 x}

Simplify

\frac{1}{ln ( 5 )} \cdot \frac{1}{x ^{4} 2 x ^{2} + 9 x} \cdot \frac{20 x ^{5} + 81 x ^{4}}{2 2 x ^{2} + 9 x} : \frac{20 x + 81}{2 ln ( 5 ) x ( 2 x + 9 )}

= \frac{20 x + 81}{2 ln ( 5 ) x ( 2 x + 9 )}

x \to 4 lim (x^{2} + x + 5)

x \to 0 lim (1 + 3 e^{x})

x \to 4 + lim (\frac{x + 4}{x - 4})

y = (\frac{x}{7} + 6)^{\frac{3}{2}}

\int \frac{2 e ^{x}}{e ^{2 x} + 2 e ^{x} + 1} d x

What is the derivative of log_{5}(x^4sqrt(2x^2+9x)) ?
The derivative of log_{5}(x^4sqrt(2x^2+9x)) is (20x+81)/(2ln(5)x(2x+9))
What is the first derivative of log_{5}(x^4sqrt(2x^2+9x)) ?
The first derivative of log_{5}(x^4sqrt(2x^2+9x)) is (20x+81)/(2ln(5)x(2x+9))