What is the derivative of log_{10}(2x-1) ?

The derivative of log_{10}(2x-1) is 2/(ln(10)(2x-1))

What is the first derivative of log_{10}(2x-1) ?

The first derivative of log_{10}(2x-1) is 2/(ln(10)(2x-1))

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\frac{d}{d x} (lo g_{10} (2 x - 1))

\frac{2}{ln ( 10 ) ( 2 x - 1 )}

Solution steps

\frac{d}{d x} (lo g_{10} (2 x - 1))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( 2 x - 1 )}{ln ( 10 )})

Take the constant out:

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

= \frac{1}{ln ( 10 )} \frac{d}{d x} (ln (2 x - 1))

Apply the chain rule:

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

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

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

= \frac{1}{ln ( 10 )} \cdot \frac{1}{2 x - 1} \cdot 2

Simplify

\frac{1}{ln ( 10 )} \cdot \frac{1}{2 x - 1} \cdot 2 : \frac{2}{ln ( 10 ) ( 2 x - 1 )}

= \frac{2}{ln ( 10 ) ( 2 x - 1 )}

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

\frac{d}{d x} ((5 x + 2)^{4} (2 x - 3)^{8})

y = 4 x, y = 0, x = 2

z \to 2 i lim (i z^{4} + 3 z^{2} - 10 i)

x y^{^{'}} = y^{2}

What is the derivative of log_{10}(2x-1) ?
The derivative of log_{10}(2x-1) is 2/(ln(10)(2x-1))
What is the first derivative of log_{10}(2x-1) ?
The first derivative of log_{10}(2x-1) is 2/(ln(10)(2x-1))