What is the derivative of log_{5}(x^2-7x) ?

The derivative of log_{5}(x^2-7x) is (2x-7)/(ln(5)(x^2-7x))

What is the first derivative of log_{5}(x^2-7x) ?

The first derivative of log_{5}(x^2-7x) is (2x-7)/(ln(5)(x^2-7x))

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\frac{d}{d x} (lo g_{5} (x^{2} - 7 x))

\frac{2 x - 7}{ln ( 5 ) ( x ^{2} - 7 x )}

Solution steps

\frac{d}{d x} (lo g_{5} (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 ( 5 )})

Take the constant out:

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

= \frac{1}{ln ( 5 )} \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 ( 5 )} \cdot \frac{1}{x ^{2} - 7 x} (2 x - 7)

Simplify

\frac{1}{ln ( 5 )} \cdot \frac{1}{x ^{2} - 7 x} (2 x - 7) : \frac{2 x - 7}{ln ( 5 ) ( x ^{2} - 7 x )}

= \frac{2 x - 7}{ln ( 5 ) ( x ^{2} - 7 x )}

v^{'} = - 0.01 v^{2}

y^{^{''}} + 2 y^{^{'}} + 2 y = sin (1 t)

x \to 5 lim (- 2)

\frac{6}{( s - 2 ) ^{4}}

f (x) = x^{2} + 8

What is the derivative of log_{5}(x^2-7x) ?
The derivative of log_{5}(x^2-7x) is (2x-7)/(ln(5)(x^2-7x))
What is the first derivative of log_{5}(x^2-7x) ?
The first derivative of log_{5}(x^2-7x) is (2x-7)/(ln(5)(x^2-7x))