What is the derivative of (x^2(x+3)/(16(x^2-4))) ?

The derivative of (x^2(x+3)/(16(x^2-4))) is (x^4-12x^2-24x)/(16(x^2-4)^2)

What is the first derivative of (x^2(x+3)/(16(x^2-4))) ?

The first derivative of (x^2(x+3)/(16(x^2-4))) is (x^4-12x^2-24x)/(16(x^2-4)^2)

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\frac{d}{d x} (\frac{x ^{2} ( x + 3 )}{16 ( x ^{2} - 4 )})

\frac{x ^{4} - 12 x ^{2} - 24 x}{16 ( x ^{2} - 4 ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{x ^{2} ( x + 3 )}{16 ( x ^{2} - 4 )})

Take the constant out:

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

= \frac{1}{16} \frac{d}{d x} (\frac{x ^{2} ( x + 3 )}{x ^{2} - 4})

Apply the Quotient Rule:

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

= \frac{1}{16} \cdot \frac{\frac{d}{d x} ( x ^{2} ( x + 3 ) ) ( x ^{2} - 4 ) - \frac{d}{d x} ( x ^{2} - 4 ) x ^{2} ( x + 3 )}{( x ^{2} - 4 ) ^{2}}

\frac{d}{d x} (x^{2} (x + 3)) = 3 x^{2} + 6 x

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

= \frac{1}{16} \cdot \frac{( 3 x ^{2} + 6 x ) ( x ^{2} - 4 ) - 2 x x ^{2} ( x + 3 )}{( x ^{2} - 4 ) ^{2}}

Simplify

\frac{1}{16} \cdot \frac{( 3 x ^{2} + 6 x ) ( x ^{2} - 4 ) - 2 x x ^{2} ( x + 3 )}{( x ^{2} - 4 ) ^{2}} : \frac{x ^{4} - 12 x ^{2} - 24 x}{16 ( x ^{2} - 4 ) ^{2}}

= \frac{x ^{4} - 12 x ^{2} - 24 x}{16 ( x ^{2} - 4 ) ^{2}}

\frac{d y}{d x} = y, y (11) = 3

\int \frac{1}{x ^{2} - x - 6} d x

\frac{\partial}{\partial x} (- \frac{y}{x})

f (x) = \frac{( x )}{x - 4}

\frac{\partial}{\partial x} (sin (x^{2} + 5))

What is the derivative of (x^2(x+3)/(16(x^2-4))) ?
The derivative of (x^2(x+3)/(16(x^2-4))) is (x^4-12x^2-24x)/(16(x^2-4)^2)
What is the first derivative of (x^2(x+3)/(16(x^2-4))) ?
The first derivative of (x^2(x+3)/(16(x^2-4))) is (x^4-12x^2-24x)/(16(x^2-4)^2)