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

The ((-ln(1+6x))/x)^' is -(6x-ln(1+6x)(1+6x))/(x^2(1+6x))

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

The first ((-ln(1+6x))/x)^' is -(6x-ln(1+6x)(1+6x))/(x^2(1+6x))

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

- \frac{6 x - ln ( 1 + 6 x ) ( 1 + 6 x )}{x ^{2} ( 1 + 6 x )}

Solution steps

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

Simplify

\frac{- ln ( 1 + 6 x )}{x} : - \frac{ln ( 1 + 6 x )}{x}

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

Take the constant out:

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

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

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= - \frac{( ln ( 1 + 6 x ) ) ^{^{'}} x - x ^{^{'}} ln ( 1 + 6 x )}{x ^{2}}

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

x^{'} = 1

= - \frac{\frac{6}{1 + 6 x} x - 1 \cdot ln ( 1 + 6 x )}{x ^{2}}

Simplify

- \frac{\frac{6}{1 + 6 x} x - 1 \cdot ln ( 1 + 6 x )}{x ^{2}} : - \frac{6 x - ln ( 1 + 6 x ) ( 1 + 6 x )}{x ^{2} ( 1 + 6 x )}

= - \frac{6 x - ln ( 1 + 6 x ) ( 1 + 6 x )}{x ^{2} ( 1 + 6 x )}

\int (2 x - 1) e^{x} d x

f (x) = x^{2} - 2 x + 3

f (x) = 9 x - 6 x, at x = 1

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

What is the ((-ln(1+6x))/x)^' ?
The ((-ln(1+6x))/x)^' is -(6x-ln(1+6x)(1+6x))/(x^2(1+6x))
What is the first ((-ln(1+6x))/x)^' ?
The first ((-ln(1+6x))/x)^' is -(6x-ln(1+6x)(1+6x))/(x^2(1+6x))