What is the derivative of log_{3}(2x+5) ?

The derivative of log_{3}(2x+5) is 2/(ln(3)(2x+5))

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

The first derivative of log_{3}(2x+5) is 2/(ln(3)(2x+5))

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derivative of

lo g_{3} (2 x + 5)

\frac{2}{ln ( 3 ) ( 2 x + 5 )}

Solution steps

\frac{d}{d x} (lo g_{3} (2 x + 5))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( 2 x + 5 )}{ln ( 3 )})

Take the constant out:

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

= \frac{1}{ln ( 3 )} \frac{d}{d x} (ln (2 x + 5))

Apply the chain rule:

\frac{1}{2 x + 5} \frac{d}{d x} (2 x + 5)

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

\frac{d}{d x} (2 x + 5) = 2

= \frac{1}{ln ( 3 )} \cdot \frac{1}{2 x + 5} \cdot 2

Simplify

\frac{1}{ln ( 3 )} \cdot \frac{1}{2 x + 5} \cdot 2 : \frac{2}{ln ( 3 ) ( 2 x + 5 )}

= \frac{2}{ln ( 3 ) ( 2 x + 5 )}

\int \frac{sec ^{2} ( t )}{1 + tan ( t )} d t

\int (\frac{2 y ^{\frac{1}{2}}}{y ^{3}}) d^{0} x

\frac{d}{d x} (\frac{x ^{2} + 1}{x ^{2} - 4})

(1 + ln (x) + \frac{y}{x}) d x + (- 1 + ln (x)) d y = 0

y = e^{3 x} + 10 e^{2 x}

What is the derivative of log_{3}(2x+5) ?
The derivative of log_{3}(2x+5) is 2/(ln(3)(2x+5))
What is the first derivative of log_{3}(2x+5) ?
The first derivative of log_{3}(2x+5) is 2/(ln(3)(2x+5))