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

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

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

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

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derivative of

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

sin^{2} (2 x + 5)

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

\frac{d}{d t} (10 + 5 t + 2 t^{2})

\frac{d}{d n} ((1 - i)^{n - 1})

(\frac{x}{x + 1}) cos (x)

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