What is the derivative of 1/(log_{4(x)}) ?

The derivative of 1/(log_{4(x)}) is -1/(2ln(2)xlog_{4)(x)^2}

What is the first derivative of 1/(log_{4(x)}) ?

The first derivative of 1/(log_{4(x)}) is -1/(2ln(2)xlog_{4)(x)^2}

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\frac{d}{d x} (\frac{1}{lo g _{4} ( x )})

- \frac{1}{2 ln ( 2 ) x lo g _{4} ( x ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{1}{lo g _{4} ( x )})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= \frac{d}{d x} (lo g_{4} (x)^{- 1})

Apply the chain rule:

- \frac{1}{( lo g _{4} ( x ) ) ^{2}} \frac{d}{d x} (lo g_{4} (x))

= - \frac{1}{( lo g _{4} ( x ) ) ^{2}} \frac{d}{d x} (lo g_{4} (x))

\frac{d}{d x} (lo g_{4} (x)) = \frac{1}{2 ln ( 2 ) x}

= - \frac{1}{( lo g _{4} ( x ) ) ^{2}} \cdot \frac{1}{2 ln ( 2 ) x}

Simplify

- \frac{1}{lo g _{4} ( x ) ^{2}} \cdot \frac{1}{2 ln ( 2 ) x} : - \frac{1}{2 ln ( 2 ) x lo g _{4} ( x ) ^{2}}

= - \frac{1}{2 ln ( 2 ) x lo g _{4} ( x ) ^{2}}

y^{^{''}} + y^{^{'}} - 20 y = 0

\int_{0}^{x} sin (t^{2}) d t

\frac{d}{d x} (- 2 x^{2} - \frac{1}{2} x + 10)

y^{^{'}} + (1 + \frac{1}{t}) y = 1

y = 6 + 5 x

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