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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derivative of

\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}}

y = \frac{x}{x + 2}, (4, \frac{2}{6})

- 2 x^{5} + 3 x^{2} - 7

\frac{\partial}{\partial x} (4 - 2 y + 4 x y)

\int (x + \frac{1}{x}) d x

\frac{d ^{2}}{d x ^{2}} (\frac{3 x - 2}{2 x + 5})

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)