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

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

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

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

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\frac{d}{d 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}}

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

\int 3 x^{\frac{1}{3}} d x

x - tan (x)

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

y^{^{''}} + y^{^{'}} + 20 y = 0

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