What is the derivative of (3x/(4x^2+9)) ?

The derivative of (3x/(4x^2+9)) is (3(-4x^2+9))/((4x^2+9)^2)

What is the first derivative of (3x/(4x^2+9)) ?

The first derivative of (3x/(4x^2+9)) is (3(-4x^2+9))/((4x^2+9)^2)

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\frac{d}{d x} (\frac{3 x}{4 x ^{2} + 9})

\frac{3 ( - 4 x ^{2} + 9 )}{( 4 x ^{2} + 9 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{3 x}{4 x ^{2} + 9})

Take the constant out:

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

= 3 \frac{d}{d x} (\frac{x}{4 x ^{2} + 9})

Apply the Quotient Rule:

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

= 3 \cdot \frac{\frac{d x}{d x} ( 4 x ^{2} + 9 ) - \frac{d}{d x} ( 4 x ^{2} + 9 ) x}{( 4 x ^{2} + 9 ) ^{2}}

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

\frac{d}{d x} (4 x^{2} + 9) = 8 x

= 3 \cdot \frac{1 \cdot ( 4 x ^{2} + 9 ) - 8 xx}{( 4 x ^{2} + 9 ) ^{2}}

Simplify

3 \cdot \frac{1 \cdot ( 4 x ^{2} + 9 ) - 8 xx}{( 4 x ^{2} + 9 ) ^{2}} : \frac{3 ( - 4 x ^{2} + 9 )}{( 4 x ^{2} + 9 ) ^{2}}

= \frac{3 ( - 4 x ^{2} + 9 )}{( 4 x ^{2} + 9 ) ^{2}}

x^{2} - 17, 16 x

y^{^{'}} = 12 - \frac{y}{30}

\frac{d}{d x} (\frac{2 x}{sin ( x )})

\frac{b ( - 12 b - 290 )}{8 - b}

\int cos (\frac{nπ x}{3}) d x

What is the derivative of (3x/(4x^2+9)) ?
The derivative of (3x/(4x^2+9)) is (3(-4x^2+9))/((4x^2+9)^2)
What is the first derivative of (3x/(4x^2+9)) ?
The first derivative of (3x/(4x^2+9)) is (3(-4x^2+9))/((4x^2+9)^2)