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

The derivative of f(x)=(x^2+2x-3)/((x+1)^2) is 8/((x+1)^3)

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

The first derivative of f(x)=(x^2+2x-3)/((x+1)^2) is 8/((x+1)^3)

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derivative of

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

\frac{8}{( x + 1 ) ^{3}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{( 2 x + 2 ) ( x + 1 ) ^{2} - 2 ( x + 1 ) ( x ^{2} + 2 x - 3 )}{( ( x + 1 ) ^{2} ) ^{2}} : \frac{8}{( x + 1 ) ^{3}}

= \frac{8}{( x + 1 ) ^{3}}

\int_{- \infty}^{0} \frac{1}{6 - 4 x} d x

y^{^{''}} + 7 y^{^{'}} + 14 y = 1096 e^{2 t} cos (8 t)

x \to 4 - lim (\frac{x}{x ^{2} - 16})

\frac{d y}{d x} = \frac{x}{25 y}, y (5) = 1

\frac{d}{d x} (9 x + 1)

What is the derivative of f(x)=(x^2+2x-3)/((x+1)^2) ?
The derivative of f(x)=(x^2+2x-3)/((x+1)^2) is 8/((x+1)^3)
What is the first derivative of f(x)=(x^2+2x-3)/((x+1)^2) ?
The first derivative of f(x)=(x^2+2x-3)/((x+1)^2) is 8/((x+1)^3)