What is the derivative of g(x)=(9ln(x))/(4x+5) ?

The derivative of g(x)=(9ln(x))/(4x+5) is (9(4x+5-4xln(x)))/(x(4x+5)^2)

What is the first derivative of g(x)=(9ln(x))/(4x+5) ?

The first derivative of g(x)=(9ln(x))/(4x+5) is (9(4x+5-4xln(x)))/(x(4x+5)^2)

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derivative of

g (x) = \frac{9 ln ( x )}{4 x + 5}

\frac{9 ( 4 x + 5 - 4 x ln ( x ) )}{x ( 4 x + 5 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{9 ln ( x )}{4 x + 5})

Take the constant out:

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

= 9 \frac{d}{d x} (\frac{ln ( x )}{4 x + 5})

Apply the Quotient Rule:

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

= 9 \cdot \frac{\frac{d}{d x} ( ln ( x ) ) ( 4 x + 5 ) - \frac{d}{d x} ( 4 x + 5 ) ln ( x )}{( 4 x + 5 ) ^{2}}

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

\frac{d}{d x} (4 x + 5) = 4

= 9 \cdot \frac{\frac{1}{x} ( 4 x + 5 ) - 4 ln ( x )}{( 4 x + 5 ) ^{2}}

Simplify

9 \cdot \frac{\frac{1}{x} ( 4 x + 5 ) - 4 ln ( x )}{( 4 x + 5 ) ^{2}} : \frac{9 ( 4 x + 5 - 4 x ln ( x ) )}{x ( 4 x + 5 ) ^{2}}

= \frac{9 ( 4 x + 5 - 4 x ln ( x ) )}{x ( 4 x + 5 ) ^{2}}

\int \frac{1}{( 2 x + 1 ) ^{3}} d x

y = 10 csc (9 x^{4} - 9 x + 4)

\frac{1}{s [ ( s + 2 ) ^{2} + 1 ]}

y^{^{'}} = \frac{x + 2 y - 1}{x + 2 y + 7}

f (x) = 8 x^{3} + 7 x

What is the derivative of g(x)=(9ln(x))/(4x+5) ?
The derivative of g(x)=(9ln(x))/(4x+5) is (9(4x+5-4xln(x)))/(x(4x+5)^2)
What is the first derivative of g(x)=(9ln(x))/(4x+5) ?
The first derivative of g(x)=(9ln(x))/(4x+5) is (9(4x+5-4xln(x)))/(x(4x+5)^2)