What is the derivative of log_{12}(x^2+nx) ?

The derivative of log_{12}(x^2+nx) is (2x+n)/(ln(12)(x^2+nx))

What is the first derivative of log_{12}(x^2+nx) ?

The first derivative of log_{12}(x^2+nx) is (2x+n)/(ln(12)(x^2+nx))

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\frac{d}{d x} (lo g_{12} (x^{2} + n x))

\frac{2 x + n}{ln ( 12 ) ( x ^{2} + n x )}

Solution steps

\frac{d}{d x} (lo g_{12} (x^{2} + n x))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( x ^{2} + n x )}{ln ( 12 )})

Take the constant out:

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

= \frac{1}{ln ( 12 )} \frac{d}{d x} (ln (x^{2} + n x))

Apply the chain rule:

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

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

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

= \frac{1}{ln ( 12 )} \cdot \frac{1}{x ^{2} + n x} (2 x + n)

Simplify

\frac{1}{ln ( 12 )} \cdot \frac{1}{x ^{2} + n x} (2 x + n) : \frac{2 x + n}{ln ( 12 ) ( x ^{2} + n x )}

= \frac{2 x + n}{ln ( 12 ) ( x ^{2} + n x )}

\frac{d}{d x} (2 sinh (\frac{x}{5}))

x \to 1 lim (\frac{x}{x ^{2} - 1})

\frac{d}{d x} ((5 x - 2 x^{4})^{3})

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

x \to 5 + lim (25 - x^{2})

What is the derivative of log_{12}(x^2+nx) ?
The derivative of log_{12}(x^2+nx) is (2x+n)/(ln(12)(x^2+nx))
What is the first derivative of log_{12}(x^2+nx) ?
The first derivative of log_{12}(x^2+nx) is (2x+n)/(ln(12)(x^2+nx))