What is the derivative of y= 7/(ln(x)) ?

The derivative of y= 7/(ln(x)) is -7/(xln^2(x))

What is the first derivative of y= 7/(ln(x)) ?

The first derivative of y= 7/(ln(x)) is -7/(xln^2(x))

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derivative of

y = \frac{7}{ln ( x )}

- \frac{7}{x ln ^{2} ( x )}

Solution steps

\frac{d}{d x} (\frac{7}{ln ( x )})

Take the constant out:

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

= 7 \frac{d}{d x} (\frac{1}{ln ( x )})

Apply exponent rule:

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

= 7 \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}

= 7 (- \frac{1}{( ln ( x ) ) ^{2}} \cdot \frac{1}{x})

Simplify

7 (- \frac{1}{ln ^{2} ( x )} \cdot \frac{1}{x}) : - \frac{7}{x ln ^{2} ( x )}

= - \frac{7}{x ln ^{2} ( x )}

\frac{d}{d x} (\frac{5 x}{6})

x \to 9 - lim (\frac{1}{x - 9})

x \to 0 + lim (x [ln (x)]^{2})

\frac{d}{d x} (2^{5 x} 3^{4 x^{2}})

\frac{d y}{d x} = \frac{x - 4}{y + 5}

What is the derivative of y= 7/(ln(x)) ?
The derivative of y= 7/(ln(x)) is -7/(xln^2(x))
What is the first derivative of y= 7/(ln(x)) ?
The first derivative of y= 7/(ln(x)) is -7/(xln^2(x))