What is the derivative of log_{4}(x^2-7x) ?

The derivative of log_{4}(x^2-7x) is (2x-7)/(2ln(2)(x^2-7x))

What is the first derivative of log_{4}(x^2-7x) ?

The first derivative of log_{4}(x^2-7x) is (2x-7)/(2ln(2)(x^2-7x))

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\frac{d}{d x} (lo g_{4} (x^{2} - 7 x))

\frac{2 x - 7}{2 ln ( 2 ) ( x ^{2} - 7 x )}

Solution steps

\frac{d}{d x} (lo g_{4} (x^{2} - 7 x))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( x ^{2} - 7 x )}{ln ( 4 )})

Take the constant out:

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

= \frac{1}{ln ( 4 )} \frac{d}{d x} (ln (x^{2} - 7 x))

Apply the chain rule:

\frac{1}{x ^{2} - 7 x} \frac{d}{d x} (x^{2} - 7 x)

= \frac{1}{x ^{2} - 7 x} \frac{d}{d x} (x^{2} - 7 x)

\frac{d}{d x} (x^{2} - 7 x) = 2 x - 7

= \frac{1}{ln ( 4 )} \cdot \frac{1}{x ^{2} - 7 x} (2 x - 7)

Simplify

\frac{1}{ln ( 4 )} \cdot \frac{1}{x ^{2} - 7 x} (2 x - 7) : \frac{2 x - 7}{2 ln ( 2 ) ( x ^{2} - 7 x )}

= \frac{2 x - 7}{2 ln ( 2 ) ( x ^{2} - 7 x )}

\int x^{2} + 9 d x

y^{^{'}} - 2 t y = 9 t^{2} e^{t^{2}}

\frac{\partial}{\partial x} (x sin (x^{2} y))

y^{8} - \frac{x ^{4} y ^{9} + 4}{x ^{5} + 4 y ^{^{'}}} = 0

\int_{0}^{1} 1 + 4 x^{2} d x

What is the derivative of log_{4}(x^2-7x) ?
The derivative of log_{4}(x^2-7x) is (2x-7)/(2ln(2)(x^2-7x))
What is the first derivative of log_{4}(x^2-7x) ?
The first derivative of log_{4}(x^2-7x) is (2x-7)/(2ln(2)(x^2-7x))