What is the derivative of y=(ln(9x))/(9x) ?

The derivative of y=(ln(9x))/(9x) is (1-ln(9x))/(9x^2)

What is the first derivative of y=(ln(9x))/(9x) ?

The first derivative of y=(ln(9x))/(9x) is (1-ln(9x))/(9x^2)

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derivative of

y = \frac{ln ( 9 x )}{9 x}

\frac{1 - ln ( 9 x )}{9 x ^{2}}

Solution steps

\frac{d}{d x} (\frac{ln ( 9 x )}{9 x})

Take the constant out:

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

= \frac{1}{9} \frac{d}{d x} (\frac{ln ( 9 x )}{x})

Apply the Quotient Rule:

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

= \frac{1}{9} \cdot \frac{\frac{d}{d x} ( ln ( 9 x ) ) x - \frac{d x}{d x} ln ( 9 x )}{x ^{2}}

\frac{d}{d x} (ln (9 x)) = \frac{1}{x}

\frac{d x}{d x} = 1

= \frac{1}{9} \cdot \frac{\frac{1}{x} x - 1 \cdot ln ( 9 x )}{x ^{2}}

Simplify

\frac{1}{9} \cdot \frac{\frac{1}{x} x - 1 \cdot ln ( 9 x )}{x ^{2}} : \frac{1 - ln ( 9 x )}{9 x ^{2}}

= \frac{1 - ln ( 9 x )}{9 x ^{2}}

\int_{- 1}^{7} (4 x + 7) - (x^{2} - 2 x) d x

\int 8 t d t

f^{^{'}} (x) = 3 e^{2 x^{2}}

y^{^{'}} = \frac{1 - 4 y ^{2}}{3 x y}

y = - 6 cos (- 2 x)

What is the derivative of y=(ln(9x))/(9x) ?
The derivative of y=(ln(9x))/(9x) is (1-ln(9x))/(9x^2)
What is the first derivative of y=(ln(9x))/(9x) ?
The first derivative of y=(ln(9x))/(9x) is (1-ln(9x))/(9x^2)