What is the derivative of f(x)=log_{41}(2x-4) ?

The derivative of f(x)=log_{41}(2x-4) is 1/(ln(41)(x-2))

What is the first derivative of f(x)=log_{41}(2x-4) ?

The first derivative of f(x)=log_{41}(2x-4) is 1/(ln(41)(x-2))

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derivative of

f (x) = lo g_{41} (2 x - 4)

\frac{1}{ln ( 41 ) ( x - 2 )}

Solution steps

\frac{d}{d x} (lo g_{41} (2 x - 4))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( 2 x - 4 )}{ln ( 41 )})

Take the constant out:

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

= \frac{1}{ln ( 41 )} \frac{d}{d x} (ln (2 x - 4))

Apply the chain rule:

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

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

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

= \frac{1}{ln ( 41 )} \cdot \frac{1}{2 x - 4} \cdot 2

Simplify

\frac{1}{ln ( 41 )} \cdot \frac{1}{2 x - 4} \cdot 2 : \frac{1}{ln ( 41 ) ( x - 2 )}

= \frac{1}{ln ( 41 ) ( x - 2 )}

\frac{d}{d x} (ln (x) cos (x))

x^{2} y^{^{''}} + 3 x y^{^{'}} - 15 y = 0

\frac{6}{s ^{2} + 5 s + 6}

x \to 4 lim (\frac{x ^{2} + x}{x + 1})

y = 17 - x^{2}, x = 0, y = 0

What is the derivative of f(x)=log_{41}(2x-4) ?
The derivative of f(x)=log_{41}(2x-4) is 1/(ln(41)(x-2))
What is the first derivative of f(x)=log_{41}(2x-4) ?
The first derivative of f(x)=log_{41}(2x-4) is 1/(ln(41)(x-2))