What is the derivative of y=(13)/(ln(x)) ?

The derivative of y=(13)/(ln(x)) is -(13)/(xln^2(x))

What is the first derivative of y=(13)/(ln(x)) ?

The first derivative of y=(13)/(ln(x)) is -(13)/(xln^2(x))

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derivative of

y = \frac{13}{ln ( x )}

- \frac{13}{x ln ^{2} ( x )}

Solution steps

\frac{d}{d x} (\frac{13}{ln ( x )})

Take the constant out:

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

= 13 \frac{d}{d x} (\frac{1}{ln ( x )})

Apply exponent rule:

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

= 13 \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}

= 13 (- \frac{1}{( ln ( x ) ) ^{2}} \cdot \frac{1}{x})

Simplify

13 (- \frac{1}{ln ^{2} ( x )} \cdot \frac{1}{x}) : - \frac{13}{x ln ^{2} ( x )}

= - \frac{13}{x ln ^{2} ( x )}

\int sin (t) cos^{2} (t) d t

x \to 0 lim (x^{3} + 2 x^{2} - 4 x + 4)

f (x) = 7 e^{0.03 x}, at x = 9

3 sin (x), 8 cos (x)

x \to 2 lim ((1 + x)^{\frac{1}{x}})

What is the derivative of y=(13)/(ln(x)) ?
The derivative of y=(13)/(ln(x)) is -(13)/(xln^2(x))
What is the first derivative of y=(13)/(ln(x)) ?
The first derivative of y=(13)/(ln(x)) is -(13)/(xln^2(x))