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

The derivative of 1/(log_{e(x)}) is -1/(xln^2(x))

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

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

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

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

Solution steps

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (lo g_{e} (x)) = \frac{1}{x}

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

Simplify

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

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

\frac{d y}{d x} + 3 y = e^{(- 5 x)}

\int \frac{3 x ^{3}}{x ^{3} - 4 x} d x

(6 n - 1) (- 2 x)^{n}

\frac{d y}{d x} = k y^{2}

f (x) = 2 x, (8, 4)

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