What is the derivative of f(x)=log_{2}(1-3x) ?

The derivative of f(x)=log_{2}(1-3x) is -3/(ln(2)(1-3x))

What is the first derivative of f(x)=log_{2}(1-3x) ?

The first derivative of f(x)=log_{2}(1-3x) is -3/(ln(2)(1-3x))

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derivative of

f (x) = lo g_{2} (1 - 3 x)

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

Solution steps

\frac{d}{d x} (lo g_{2} (1 - 3 x))

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

sin (2 x + π)

y^{^{''}} + 4 y^{^{'}} + 3 y = 4 e^{x}, y (0) = 0, y^{^{'}} (0) = 2

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

\frac{\partial}{\partial x} (cos (y) + y e^{x})

\int \frac{16}{x ^{2} 16 - x ^{2}} d x

What is the derivative of f(x)=log_{2}(1-3x) ?
The derivative of f(x)=log_{2}(1-3x) is -3/(ln(2)(1-3x))
What is the first derivative of f(x)=log_{2}(1-3x) ?
The first derivative of f(x)=log_{2}(1-3x) is -3/(ln(2)(1-3x))