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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\frac{d x}{d t} (1) - x = e^{2 t}

\frac{\partial}{\partial x} (2 x^{2} + y^{2} - 9 y)

2 y^{^{''}} + 7 y^{^{'}} + 5 y = 0

\int_{3}^{5} (- x^{2} + 8 x - 12) d x

x \to 2 lim (3 x + 6)

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