What is the derivative of log_{5}(3x^2+4x) ?

The derivative of log_{5}(3x^2+4x) is (6x+4)/(ln(5)(3x^2+4x))

What is the first derivative of log_{5}(3x^2+4x) ?

The first derivative of log_{5}(3x^2+4x) is (6x+4)/(ln(5)(3x^2+4x))

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\frac{d}{d x} (lo g_{5} (3 x^{2} + 4 x))

\frac{6 x + 4}{ln ( 5 ) ( 3 x ^{2} + 4 x )}

Solution steps

\frac{d}{d x} (lo g_{5} (3 x^{2} + 4 x))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( 3 x ^{2} + 4 x )}{ln ( 5 )})

Take the constant out:

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

= \frac{1}{ln ( 5 )} \frac{d}{d x} (ln (3 x^{2} + 4 x))

Apply the chain rule:

\frac{1}{3 x ^{2} + 4 x} \frac{d}{d x} (3 x^{2} + 4 x)

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

\frac{d}{d x} (3 x^{2} + 4 x) = 6 x + 4

= \frac{1}{ln ( 5 )} \cdot \frac{1}{3 x ^{2} + 4 x} (6 x + 4)

Simplify

\frac{1}{ln ( 5 )} \cdot \frac{1}{3 x ^{2} + 4 x} (6 x + 4) : \frac{6 x + 4}{ln ( 5 ) ( 3 x ^{2} + 4 x )}

= \frac{6 x + 4}{ln ( 5 ) ( 3 x ^{2} + 4 x )}

(\frac{8 x}{8 x + 6})^{7 x}

y = \frac{4}{3} π r^{6}

\int_{0}^{1} π (4 x)^{2} d x

\int_{0}^{1} e^{- x} + 3 x^{2} d x

x \to \infty + lim (\frac{1}{x !})

What is the derivative of log_{5}(3x^2+4x) ?
The derivative of log_{5}(3x^2+4x) is (6x+4)/(ln(5)(3x^2+4x))
What is the first derivative of log_{5}(3x^2+4x) ?
The first derivative of log_{5}(3x^2+4x) is (6x+4)/(ln(5)(3x^2+4x))