What is the ((x^2+x+4)/(x+1))^' ?

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

What is the first ((x^2+x+4)/(x+1))^' ?

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

(x^{2} + x + 4)^{'} = 2 x + 1

(x + 1)^{'} = 1

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

Simplify

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

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

8 t e^{2 t}

f (x) = \frac{5 - x e ^{x}}{x + e ^{x}}

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

y = x e^{- 2 x}

\frac{1}{2} sin^{2} (x)

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