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

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

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

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

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

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

Solution steps

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

Take the constant out:

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

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

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

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

\int \frac{x ^{2} - 6 x + 9}{x ^{2} - 6 x + 45} d x

y^{^{'}} = (3 x - 2)^{2} (5 x + 4)

t^{3} - 12 t

y^{^{''}} = \frac{1}{x ^{2}} y

\int_{0}^{x} sin (t) d t

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