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

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

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

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

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

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

Solution steps

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

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

x \to 0 + lim ((tan (8 x))^{x})

(x^{2} + 1)^{x}

\frac{16}{s ( s ^{2} + 4 s + 16 )}

\frac{1 - y}{x - 1}

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