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

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

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

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

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

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

Solution steps

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

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

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

Apply the Quotient Rule:

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

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

\frac{d x}{d x} = 1

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

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

Simplify

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

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

x \frac{d y}{d x} = y + x e^{x}

\frac{d}{d x} (\frac{3}{x + 5})

\int 19 tan^{3} (x) d x

\int (1 - cos (θ))^{3} d θ

x^{^{''}} + b x = 0

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