What is the (1/(ln(x)))^' ?

The (1/(ln(x)))^' is -1/(xln^2(x))

What is the first (1/(ln(x)))^' ?

The first (1/(ln(x)))^' is -1/(xln^2(x))

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(\frac{1}{ln ( x )})^{^{'}}

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

Solution steps

(\frac{1}{ln ( x )})^{'}

Apply exponent rule:

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

= (ln^{- 1} (x))^{^{'}}

Apply the chain rule:

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

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

(ln (x))^{'} = \frac{1}{x}

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

Simplify

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

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

(2 x + 3)^{^{'}}

\frac{d}{d x} (I n (x^{3} + x))

f (x) = 2 x^{2} + 7 x

x^{4} + 7 e^{x}

y = e^{3 x}, at x = \frac{1}{3} ln (5)