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

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

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

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

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

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

Solution steps

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

Apply log rule:

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

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

Take the constant out:

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

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

f (x) = x \cdot e^{- x + 1}

4 + 5 x^{2} - 2 x^{3}

\frac{d}{d x} (\frac{sin ( x ) + cos ( x )}{e ^{x}})

x \to 5 - lim (8 - x - x^{2})

y^{^{'}} = \frac{x - 2}{y - 2}

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