What is the derivative of log_{6}(x^2-5x) ?

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

What is the first derivative of log_{6}(x^2-5x) ?

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

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derivative of

lo g_{6} (x^{2} - 5 x)

\frac{2 x - 5}{ln ( 6 ) ( x ^{2} - 5 x )}

Solution steps

\frac{d}{d x} (lo g_{6} (x^{2} - 5 x))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( x ^{2} - 5 x )}{ln ( 6 )})

Take the constant out:

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

= \frac{1}{ln ( 6 )} \frac{d}{d x} (ln (x^{2} - 5 x))

Apply the chain rule:

\frac{1}{x ^{2} - 5 x} \frac{d}{d x} (x^{2} - 5 x)

= \frac{1}{x ^{2} - 5 x} \frac{d}{d x} (x^{2} - 5 x)

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

= \frac{1}{ln ( 6 )} \cdot \frac{1}{x ^{2} - 5 x} (2 x - 5)

Simplify

\frac{1}{ln ( 6 )} \cdot \frac{1}{x ^{2} - 5 x} (2 x - 5) : \frac{2 x - 5}{ln ( 6 ) ( x ^{2} - 5 x )}

= \frac{2 x - 5}{ln ( 6 ) ( x ^{2} - 5 x )}

\int_{0}^{\frac{π}{3}} 7 sec^{2} (x) d x

\frac{\partial}{\partial x} (3 ln (x))

a re a (x + 6), x, - 2, 3

\int sec^{6} (x) tan (x) d x

d er i v a t i v e y = ln (7 x)

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