What is the derivative of (2x/(9x^2+1)) ?

The derivative of (2x/(9x^2+1)) is (2(-9x^2+1))/((9x^2+1)^2)

What is the first derivative of (2x/(9x^2+1)) ?

The first derivative of (2x/(9x^2+1)) is (2(-9x^2+1))/((9x^2+1)^2)

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\frac{d}{d x} (\frac{2 x}{9 x ^{2} + 1})

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

Solution steps

\frac{d}{d x} (\frac{2 x}{9 x ^{2} + 1})

Take the constant out:

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

= 2 \frac{d}{d x} (\frac{x}{9 x ^{2} + 1})

Apply the Quotient Rule:

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

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

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

\frac{d}{d x} (9 x^{2} + 1) = 18 x

= 2 \cdot \frac{1 \cdot ( 9 x ^{2} + 1 ) - 18 xx}{( 9 x ^{2} + 1 ) ^{2}}

Simplify

2 \cdot \frac{1 \cdot ( 9 x ^{2} + 1 ) - 18 xx}{( 9 x ^{2} + 1 ) ^{2}} : \frac{2 ( - 9 x ^{2} + 1 )}{( 9 x ^{2} + 1 ) ^{2}}

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

\int \frac{1}{4 x ^{2} + 2 x + 1} d x

\frac{1}{u}

\int_{0}^{0.5} tan (x) d x

\int [(x + 3) (x - 4) + \frac{1}{x ^{2}}] d x

\int \frac{2 x ^{3} + x ^{2} - 6 x + 7}{x ^{2} + x - 6} d x

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