What is the first (-ln(1+6x))^' ?

The first (-ln(1+6x))^' is -6/(1+6x)

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(-ln(1+6x))^'

(- ln (1 + 6 x))^{^{'}}

- \frac{6}{1 + 6 x}

Solution steps

(- ln (1 + 6 x))^{'}

Take the constant out:

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

= - (ln (1 + 6 x))^{^{'}}

Apply the chain rule:

\frac{1}{1 + 6 x} (1 + 6 x)^{'}

= \frac{1}{1 + 6 x} (1 + 6 x)^{'}

(1 + 6 x)^{'} = 6

= - \frac{1}{1 + 6 x} \cdot 6

Simplify

- \frac{1}{1 + 6 x} \cdot 6 : - \frac{6}{1 + 6 x}

= - \frac{6}{1 + 6 x}

x \to - 5 lim (6 x - \frac{2}{x ^{3}})

x - 4

x \to 0 lim (\frac{1}{x} - \frac{1}{x ^{2}})

x \to - \infty lim (\frac{4}{x} - 5)

\frac{d y}{d x} = \frac{x ^{3} + y ^{3}}{3 x y ^{2}}