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

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

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

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

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

x^{'} = 1

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

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

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

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

\int_{0}^{2 π} 1 d θ

4 arcsin (e^{\frac{x}{2}})

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

\frac{\partial}{\partial x} (sin (- x) cos (2 y))

x \to 0 lim (\frac{ln ( 3 x + 1 )}{x})

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