What is the derivative of log_{4}(2x+3) ?

The derivative of log_{4}(2x+3) is 1/(ln(2)(2x+3))

What is the first derivative of log_{4}(2x+3) ?

The first derivative of log_{4}(2x+3) is 1/(ln(2)(2x+3))

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\frac{d}{d x} (lo g_{4} (2 x + 3))

\frac{1}{ln ( 2 ) ( 2 x + 3 )}

Solution steps

\frac{d}{d x} (lo g_{4} (2 x + 3))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( 2 x + 3 )}{ln ( 4 )})

Take the constant out:

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

= \frac{1}{ln ( 4 )} \frac{d}{d x} (ln (2 x + 3))

Apply the chain rule:

\frac{1}{2 x + 3} \frac{d}{d x} (2 x + 3)

= \frac{1}{2 x + 3} \frac{d}{d x} (2 x + 3)

\frac{d}{d x} (2 x + 3) = 2

= \frac{1}{ln ( 4 )} \cdot \frac{1}{2 x + 3} \cdot 2

Simplify

\frac{1}{ln ( 4 )} \cdot \frac{1}{2 x + 3} \cdot 2 : \frac{1}{ln ( 2 ) ( 2 x + 3 )}

= \frac{1}{ln ( 2 ) ( 2 x + 3 )}

f (x) = (5 t^{2} - 2 t + 1)^{\frac{3}{2}}

\int_{1}^{2} \frac{1}{( x + 1 ) ^{2}} d x

f (x) = ∣ x - 1 ∣

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

\int \frac{x - 2}{x ^{2} - x} d x

What is the derivative of log_{4}(2x+3) ?
The derivative of log_{4}(2x+3) is 1/(ln(2)(2x+3))
What is the first derivative of log_{4}(2x+3) ?
The first derivative of log_{4}(2x+3) is 1/(ln(2)(2x+3))