What is the derivative of y= 4/(ln(x)) ?

The derivative of y= 4/(ln(x)) is -4/(xln^2(x))

What is the first derivative of y= 4/(ln(x)) ?

The first derivative of y= 4/(ln(x)) is -4/(xln^2(x))

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derivative of

y = \frac{4}{ln ( x )}

- \frac{4}{x ln ^{2} ( x )}

Solution steps

\frac{d}{d x} (\frac{4}{ln ( x )})

Take the constant out:

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

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

Apply exponent rule:

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

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

Apply the chain rule:

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

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

\frac{d}{d x} (ln (x)) = \frac{1}{x}

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

Simplify

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

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

\int \frac{x ^{4} + 1}{x ^{5} + 4 x ^{3}} d x

n = 1 \sum \infty (- 1)^{n - 1}

\frac{\partial}{\partial x} (2 x y - x^{3} - y^{2})

\int t (e^{- s t}) d t

\frac{3}{2} t + 2 e^{- t} - \frac{1}{4} e^{- 2 t} - \frac{7}{4}

What is the derivative of y= 4/(ln(x)) ?
The derivative of y= 4/(ln(x)) is -4/(xln^2(x))
What is the first derivative of y= 4/(ln(x)) ?
The first derivative of y= 4/(ln(x)) is -4/(xln^2(x))