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

The derivative of (x^2+4/(4-x^2)) is (16x)/((4-x^2)^2)

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

The first derivative of (x^2+4/(4-x^2)) is (16x)/((4-x^2)^2)

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

\frac{16 x}{( 4 - x ^{2} ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{2 x ( 4 - x ^{2} ) - ( - 2 x ) ( x ^{2} + 4 )}{( 4 - x ^{2} ) ^{2}} : \frac{16 x}{( 4 - x ^{2} ) ^{2}}

= \frac{16 x}{( 4 - x ^{2} ) ^{2}}

\frac{10}{( 1 - x ) ^{3}}

y^{^{'}} = y - x^{2}

(x e^{4 x})^{^{'}}

arcsin (x^{3})

f (x) = 7 - x^{3}

What is the derivative of (x^2+4/(4-x^2)) ?
The derivative of (x^2+4/(4-x^2)) is (16x)/((4-x^2)^2)
What is the first derivative of (x^2+4/(4-x^2)) ?
The first derivative of (x^2+4/(4-x^2)) is (16x)/((4-x^2)^2)