What is the derivative of (6x/(x^2+9)) ?

The derivative of (6x/(x^2+9)) is (6(-x^2+9))/((x^2+9)^2)

What is the first derivative of (6x/(x^2+9)) ?

The first derivative of (6x/(x^2+9)) is (6(-x^2+9))/((x^2+9)^2)

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\frac{d}{d x} (\frac{6 x}{x ^{2} + 9})

\frac{6 ( - x ^{2} + 9 )}{( x ^{2} + 9 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 x}{x ^{2} + 9})

Take the constant out:

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

= 6 \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}}

= 6 \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

= 6 \cdot \frac{1 \cdot ( x ^{2} + 9 ) - 2 xx}{( x ^{2} + 9 ) ^{2}}

Simplify

6 \cdot \frac{1 \cdot ( x ^{2} + 9 ) - 2 xx}{( x ^{2} + 9 ) ^{2}} : \frac{6 ( - x ^{2} + 9 )}{( x ^{2} + 9 ) ^{2}}

= \frac{6 ( - x ^{2} + 9 )}{( x ^{2} + 9 ) ^{2}}

\frac{\partial}{\partial y} (\frac{a x + b y}{c x + d y})

\frac{8}{9 t ^{6}}

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

x \to - 1 lim (x^{5})

\int \frac{1}{x ^{2} - 100} d x

What is the derivative of (6x/(x^2+9)) ?
The derivative of (6x/(x^2+9)) is (6(-x^2+9))/((x^2+9)^2)
What is the first derivative of (6x/(x^2+9)) ?
The first derivative of (6x/(x^2+9)) is (6(-x^2+9))/((x^2+9)^2)