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

The (((x+1))/((x+1)^2+4))^' is (-x^2-2x+3)/((x^2+2x+5)^2)

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

The first (((x+1))/((x+1)^2+4))^' is (-x^2-2x+3)/((x^2+2x+5)^2)

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

\frac{- x ^{2} - 2 x + 3}{( x ^{2} + 2 x + 5 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

(x + 1)^{'} = 1

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

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

Simplify

\frac{1 \cdot ( ( x + 1 ) ^{2} + 4 ) - 2 ( x + 1 ) ( x + 1 )}{( ( x + 1 ) ^{2} + 4 ) ^{2}} : \frac{- x ^{2} - 2 x + 3}{( x ^{2} + 2 x + 5 ) ^{2}}

= \frac{- x ^{2} - 2 x + 3}{( x ^{2} + 2 x + 5 ) ^{2}}

\frac{d}{d x} (x^{5} sin (3 x))

\frac{d}{d x} ((x^{5} - \frac{6}{x})^{- 2})

\int_{0}^{1} x e^{x} d x

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

\frac{\partial}{\partial y} (x yz + x y^{5} + y z^{5} + z x^{5})

What is the (((x+1))/((x+1)^2+4))^' ?
The (((x+1))/((x+1)^2+4))^' is (-x^2-2x+3)/((x^2+2x+5)^2)
What is the first (((x+1))/((x+1)^2+4))^' ?
The first (((x+1))/((x+1)^2+4))^' is (-x^2-2x+3)/((x^2+2x+5)^2)