What is the derivative of f(x)=log_{4}(x^2-3x) ?

The derivative of f(x)=log_{4}(x^2-3x) is (2x-3)/(2ln(2)(x^2-3x))

What is the first derivative of f(x)=log_{4}(x^2-3x) ?

The first derivative of f(x)=log_{4}(x^2-3x) is (2x-3)/(2ln(2)(x^2-3x))

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derivative of

f (x) = lo g_{4} (x^{2} - 3 x)

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

Solution steps

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

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

\frac{d y}{d x} = e^{5 y + 6 x}, y (0) = 2

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

\frac{d y}{d x} = y (1 - 0.0002 y)

ln (x^{2} + 4)

x \to 0 lim (2 f (x) - 1)

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