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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

f (x) = 7 x - x^{2}, at x = 1

(sin (2 t))^{^{'}}

\int_{0}^{π} 5 x^{2} sin (\frac{1}{2} x) d x

x \to 1 lim (9 x + 5)

t^{2} y^{^{''}} + t y^{^{'}} + 4 y = sin (2 ln (t))

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