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

The derivative of-(6x/((x^2+9)^2)) is -(18(-x^2+3))/((x^2+9)^3)

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

The first derivative of-(6x/((x^2+9)^2)) is -(18(-x^2+3))/((x^2+9)^3)

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

- \frac{18 ( - x ^{2} + 3 )}{( x ^{2} + 9 ) ^{3}}

Solution steps

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

Take the constant out:

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

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

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

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

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

= - 6 \cdot \frac{1 \cdot ( x ^{2} + 9 ) ^{2} - 4 x ( x ^{2} + 9 ) x}{( ( x ^{2} + 9 ) ^{2} ) ^{2}}

Simplify

- 6 \cdot \frac{1 \cdot ( x ^{2} + 9 ) ^{2} - 4 x ( x ^{2} + 9 ) x}{( ( x ^{2} + 9 ) ^{2} ) ^{2}} : - \frac{18 ( - x ^{2} + 3 )}{( x ^{2} + 9 ) ^{3}}

= - \frac{18 ( - x ^{2} + 3 )}{( x ^{2} + 9 ) ^{3}}

0.7 x^{3} 5^{x}

\int sinh (d) x d x

f (x) = 5 x + 6, at x = 6

5 \cdot \frac{d Q}{d t} + \frac{Q}{3} = 0

f (x) = 5 e^{x}

What is the derivative of-(6x/((x^2+9)^2)) ?
The derivative of-(6x/((x^2+9)^2)) is -(18(-x^2+3))/((x^2+9)^3)
What is the first derivative of-(6x/((x^2+9)^2)) ?
The first derivative of-(6x/((x^2+9)^2)) is -(18(-x^2+3))/((x^2+9)^3)