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

The derivative of (6x^2/((x^3+1)^2)) is (12x(-2x^3+1))/((x^3+1)^3)

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

The first derivative of (6x^2/((x^3+1)^2)) is (12x(-2x^3+1))/((x^3+1)^3)

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

\frac{12 x ( - 2 x ^{3} + 1 )}{( x ^{3} + 1 ) ^{3}}

Solution steps

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

Take the constant out:

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

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

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

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

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

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

Simplify

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

= \frac{12 x ( - 2 x ^{3} + 1 )}{( x ^{3} + 1 ) ^{3}}

y^{^{''}} + 3 y^{^{'}} = 0, y (0) = 2, y^{^{'}} (0) = - 3

\int \frac{1}{1 - 3 y} d y

y^{^{''}} + 2 y^{^{'}} = 8 sin (2 t) + 32 cos (2 t)

f (x) = 5 x^{2} - 6

f (x) = 4 x^{2} + 2 x, (- 1, 2)

What is the derivative of (6x^2/((x^3+1)^2)) ?
The derivative of (6x^2/((x^3+1)^2)) is (12x(-2x^3+1))/((x^3+1)^3)
What is the first derivative of (6x^2/((x^3+1)^2)) ?
The first derivative of (6x^2/((x^3+1)^2)) is (12x(-2x^3+1))/((x^3+1)^3)