What is the derivative of y=log_{2}(4-9x) ?

The derivative of y=log_{2}(4-9x) is -9/(ln(2)(4-9x))

What is the first derivative of y=log_{2}(4-9x) ?

The first derivative of y=log_{2}(4-9x) is -9/(ln(2)(4-9x))

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derivative of

y = lo g_{2} (4 - 9 x)

- \frac{9}{ln ( 2 ) ( 4 - 9 x )}

Solution steps

\frac{d}{d x} (lo g_{2} (4 - 9 x))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( 4 - 9 x )}{ln ( 2 )})

Take the constant out:

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

= \frac{1}{ln ( 2 )} \frac{d}{d x} (ln (4 - 9 x))

Apply the chain rule:

\frac{1}{4 - 9 x} \frac{d}{d x} (4 - 9 x)

= \frac{1}{4 - 9 x} \frac{d}{d x} (4 - 9 x)

\frac{d}{d x} (4 - 9 x) = - 9

= \frac{1}{ln ( 2 )} \cdot \frac{1}{4 - 9 x} (- 9)

Simplify

\frac{1}{ln ( 2 )} \cdot \frac{1}{4 - 9 x} (- 9) : - \frac{9}{ln ( 2 ) ( 4 - 9 x )}

= - \frac{9}{ln ( 2 ) ( 4 - 9 x )}

\frac{2 x}{x ^{2} + 1}

\int x^{5} 1 + x^{6} d x

x \to - 1 + lim (\frac{x}{1 - x ^{4}})

\frac{\partial}{\partial x} (x^{0.75} y^{0.25})

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

What is the derivative of y=log_{2}(4-9x) ?
The derivative of y=log_{2}(4-9x) is -9/(ln(2)(4-9x))
What is the first derivative of y=log_{2}(4-9x) ?
The first derivative of y=log_{2}(4-9x) is -9/(ln(2)(4-9x))