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

The derivative of y/(1+x^2) is (\frac{dy)/(dx)(1+x^2)-2xy}{(1+x^2)^2}

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

The first derivative of y/(1+x^2) is (\frac{dy)/(dx)(1+x^2)-2xy}{(1+x^2)^2}

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

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

\int \frac{x ^{3}}{5} d x

\int_{1}^{\infty} \frac{1}{n ^{2}}

x, 2 - x, 0

e^{- t^{2}}

x \to 0 lim (\frac{2 x - 2}{x})

What is the derivative of y/(1+x^2) ?
The derivative of y/(1+x^2) is (\frac{dy)/(dx)(1+x^2)-2xy}{(1+x^2)^2}
What is the first derivative of y/(1+x^2) ?
The first derivative of y/(1+x^2) is (\frac{dy)/(dx)(1+x^2)-2xy}{(1+x^2)^2}