What is the derivative of (6x)/(4+x^2) ?

The derivative of (6x)/(4+x^2) is (6(4-x^2))/((4+x^2)^2)

What is the first derivative of (6x)/(4+x^2) ?

The first derivative of (6x)/(4+x^2) is (6(4-x^2))/((4+x^2)^2)

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derivative of

\frac{6 x}{4 + x ^{2}}

\frac{6 ( 4 - x ^{2} )}{( 4 + x ^{2} ) ^{2}}

Solution steps

\frac{d}{d x} (\frac{6 x}{4 + x ^{2}})

Take the constant out:

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

= 6 \frac{d}{d x} (\frac{x}{4 + x ^{2}})

Apply the Quotient Rule:

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

= 6 \cdot \frac{\frac{d x}{d x} ( 4 + x ^{2} ) - \frac{d}{d x} ( 4 + x ^{2} ) x}{( 4 + x ^{2} ) ^{2}}

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

\frac{d}{d x} (4 + x^{2}) = 2 x

= 6 \cdot \frac{1 \cdot ( 4 + x ^{2} ) - 2 xx}{( 4 + x ^{2} ) ^{2}}

Simplify

6 \cdot \frac{1 \cdot ( 4 + x ^{2} ) - 2 xx}{( 4 + x ^{2} ) ^{2}} : \frac{6 ( 4 - x ^{2} )}{( 4 + x ^{2} ) ^{2}}

= \frac{6 ( 4 - x ^{2} )}{( 4 + x ^{2} ) ^{2}}

\frac{\partial}{\partial x} (cos (6 x))

\frac{d}{d x} (x^{3} + 8 x^{2} + 5 x - 50)

\int \frac{1}{2 + 8 x} d x

\int sec (13 w) tan (13 w) d w

\int_{0}^{\frac{π}{4}} tan^{3} (x) d x

What is the derivative of (6x)/(4+x^2) ?
The derivative of (6x)/(4+x^2) is (6(4-x^2))/((4+x^2)^2)
What is the first derivative of (6x)/(4+x^2) ?
The first derivative of (6x)/(4+x^2) is (6(4-x^2))/((4+x^2)^2)