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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

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

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

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

Simplify

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

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

x \to 0 lim (x^{2} - 2 x^{3})

n = 0 \sum \infty \frac{1}{n !} x^{n}

x \to 0 lim (- x^{2} + 4 x)

y^{^{''}} - \frac{3}{x} y^{^{'}} + \frac{4}{x ^{2}} y = 0

x \to 2 lim (\frac{( ∣ x - 2 ∣ )}{2})

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