What is the derivative of log_{5}(3x^2-7x) ?

The derivative of log_{5}(3x^2-7x) is (6x-7)/(ln(5)(3x^2-7x))

What is the first derivative of log_{5}(3x^2-7x) ?

The first derivative of log_{5}(3x^2-7x) is (6x-7)/(ln(5)(3x^2-7x))

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\frac{d}{d x} (lo g_{5} (3 x^{2} - 7 x))

\frac{6 x - 7}{ln ( 5 ) ( 3 x ^{2} - 7 x )}

Solution steps

\frac{d}{d x} (lo g_{5} (3 x^{2} - 7 x))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( 3 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 (3 x^{2} - 7 x))

Apply the chain rule:

\frac{1}{3 x ^{2} - 7 x} \frac{d}{d x} (3 x^{2} - 7 x)

= \frac{1}{3 x ^{2} - 7 x} \frac{d}{d x} (3 x^{2} - 7 x)

\frac{d}{d x} (3 x^{2} - 7 x) = 6 x - 7

= \frac{1}{ln ( 5 )} \cdot \frac{1}{3 x ^{2} - 7 x} (6 x - 7)

Simplify

\frac{1}{ln ( 5 )} \cdot \frac{1}{3 x ^{2} - 7 x} (6 x - 7) : \frac{6 x - 7}{ln ( 5 ) ( 3 x ^{2} - 7 x )}

= \frac{6 x - 7}{ln ( 5 ) ( 3 x ^{2} - 7 x )}

\int \frac{7 x + 29}{x ^{2} + 8 x + 15} d x

f (x) = x^{2} - 2 x + 4

\frac{\partial}{\partial x} (x ln (z^{7} + x^{7}))

x (x^{4} + 2)^{3}

sec (θ) tan (θ)

What is the derivative of log_{5}(3x^2-7x) ?
The derivative of log_{5}(3x^2-7x) is (6x-7)/(ln(5)(3x^2-7x))
What is the first derivative of log_{5}(3x^2-7x) ?
The first derivative of log_{5}(3x^2-7x) is (6x-7)/(ln(5)(3x^2-7x))