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

The derivative of log_{4}(x^2-9x) is (2x-9)/(2ln(2)(x^2-9x))

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

The first derivative of log_{4}(x^2-9x) is (2x-9)/(2ln(2)(x^2-9x))

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

\frac{2 x - 9}{2 ln ( 2 ) ( x ^{2} - 9 x )}

Solution steps

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

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (x^{2} - 9 x) = 2 x - 9

= \frac{1}{ln ( 4 )} \cdot \frac{1}{x ^{2} - 9 x} (2 x - 9)

Simplify

\frac{1}{ln ( 4 )} \cdot \frac{1}{x ^{2} - 9 x} (2 x - 9) : \frac{2 x - 9}{2 ln ( 2 ) ( x ^{2} - 9 x )}

= \frac{2 x - 9}{2 ln ( 2 ) ( x ^{2} - 9 x )}

\int \frac{3}{- 1 - x} d x

x \to 0 lim (\frac{1}{2} + x^{2})

\int 9 x^{2} sin (π x) d x

\frac{d}{d t} (3 t^{3})

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

What is the derivative of log_{4}(x^2-9x) ?
The derivative of log_{4}(x^2-9x) is (2x-9)/(2ln(2)(x^2-9x))
What is the first derivative of log_{4}(x^2-9x) ?
The first derivative of log_{4}(x^2-9x) is (2x-9)/(2ln(2)(x^2-9x))