What is the derivative of log_{5}(arccos(x^3-2)) ?

The derivative of log_{5}(arccos(x^3-2)) is -(3x^2)/(ln(5)arccos(x^3-2)sqrt(-x^6+4x^3-3))

What is the first derivative of log_{5}(arccos(x^3-2)) ?

The first derivative of log_{5}(arccos(x^3-2)) is -(3x^2)/(ln(5)arccos(x^3-2)sqrt(-x^6+4x^3-3))

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\frac{d}{d x} (lo g_{5} (arccos (x^{3} - 2)))

- \frac{3 x ^{2}}{ln ( 5 ) arccos ( x ^{3} - 2 ) - x ^{6} + 4 x ^{3} - 3}

Solution steps

\frac{d}{d x} (lo g_{5} (arccos (x^{3} - 2)))

Apply log rule:

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

= \frac{d}{d x} (\frac{ln ( arccos ( x ^{3} - 2 ) )}{ln ( 5 )})

Take the constant out:

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

= \frac{1}{ln ( 5 )} \frac{d}{d x} (ln (arccos (x^{3} - 2)))

Apply the chain rule:

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

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

\frac{d}{d x} (arccos (x^{3} - 2)) = - \frac{3 x ^{2}}{- x ^{6} + 4 x ^{3} - 3}

= \frac{1}{ln ( 5 )} \cdot \frac{1}{arccos ( x ^{3} - 2 )} (- \frac{3 x ^{2}}{- x ^{6} + 4 x ^{3} - 3})

Simplify

\frac{1}{ln ( 5 )} \cdot \frac{1}{arccos ( x ^{3} - 2 )} (- \frac{3 x ^{2}}{- x ^{6} + 4 x ^{3} - 3}) : - \frac{3 x ^{2}}{ln ( 5 ) arccos ( x ^{3} - 2 ) - x ^{6} + 4 x ^{3} - 3}

= - \frac{3 x ^{2}}{ln ( 5 ) arccos ( x ^{3} - 2 ) - x ^{6} + 4 x ^{3} - 3}

\frac{\partial}{\partial x} (- 2 e^{y} sin (- x) + 3 y - 4 x y)

\frac{\partial}{\partial x} (3 (x^{\frac{1}{3}} + y^{\frac{1}{3}}))

\int_{2}^{5} \frac{4}{x} d x

120 + x^{3}

\int 4 x^{- 5} + 6 x^{- 1} - 1 d x

What is the derivative of log_{5}(arccos(x^3-2)) ?
The derivative of log_{5}(arccos(x^3-2)) is -(3x^2)/(ln(5)arccos(x^3-2)sqrt(-x^6+4x^3-3))
What is the first derivative of log_{5}(arccos(x^3-2)) ?
The first derivative of log_{5}(arccos(x^3-2)) is -(3x^2)/(ln(5)arccos(x^3-2)sqrt(-x^6+4x^3-3))