What is the derivative of (6(x^2+1)/((x^2-3)^3)) ?

The derivative of (6(x^2+1)/((x^2-3)^3)) is -(24x(x^2+3))/((x^2-3)^4)

What is the first derivative of (6(x^2+1)/((x^2-3)^3)) ?

The first derivative of (6(x^2+1)/((x^2-3)^3)) is -(24x(x^2+3))/((x^2-3)^4)

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\frac{d}{d x} (\frac{6 ( x ^{2} + 1 )}{( x ^{2} - 3 ) ^{3}})

- \frac{24 x ( x ^{2} + 3 )}{( x ^{2} - 3 ) ^{4}}

Solution steps

\frac{d}{d x} (\frac{6 ( x ^{2} + 1 )}{( x ^{2} - 3 ) ^{3}})

Take the constant out:

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

= 6 \frac{d}{d x} (\frac{x ^{2} + 1}{( x ^{2} - 3 ) ^{3}})

Apply the Quotient Rule:

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

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

\frac{d}{d x} (x^{2} + 1) = 2 x

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

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

Simplify

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

= - \frac{24 x ( x ^{2} + 3 )}{( x ^{2} - 3 ) ^{4}}

e^{x}, x e^{x^{2}}, 0, 1

\frac{2 s + 5}{s ^{2} + 6 s + 34}

\frac{d}{d x} (x^{\frac{1}{4}} - 2)

(y + x^{2} + y^{2}) d x = x d y

\int \frac{6 x ^{2}}{( x ^{6} + 1 ) ( x ^{3} + 1 )} d x

What is the derivative of (6(x^2+1)/((x^2-3)^3)) ?
The derivative of (6(x^2+1)/((x^2-3)^3)) is -(24x(x^2+3))/((x^2-3)^4)
What is the first derivative of (6(x^2+1)/((x^2-3)^3)) ?
The first derivative of (6(x^2+1)/((x^2-3)^3)) is -(24x(x^2+3))/((x^2-3)^4)