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

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

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

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

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derivative of

y = lo g_{2} (3 x + 4)

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

Solution steps

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

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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