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

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

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

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

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\frac{d}{d x} (- \frac{2 x}{4 - x ^{2}})

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

Solution steps

\frac{d}{d x} (- \frac{2 x}{4 - x ^{2}})

Take the constant out:

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

= - 2 \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}}

= - 2 \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

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

Simplify

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

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

y^{^{''}} - 9 y = sin (2 t)

\int_{7}^{9} (x^{2} + 2 x - 2) d x

\frac{d}{d x} ((3)^{x})

\int \frac{2 x + 1}{4 x ^{2} + 4 x + 7} d x

f (x) = 3 sin (x)

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