What is the derivative of (8x/(x^2+3)) ?

The derivative of (8x/(x^2+3)) is (8(-x^2+3))/((x^2+3)^2)

What is the first derivative of (8x/(x^2+3)) ?

The first derivative of (8x/(x^2+3)) is (8(-x^2+3))/((x^2+3)^2)

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

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

Solution steps

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

Take the constant out:

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

= 8 \frac{d}{d x} (\frac{x}{x ^{2} + 3})

Apply the Quotient Rule:

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

= 8 \cdot \frac{\frac{d x}{d x} ( x ^{2} + 3 ) - \frac{d}{d x} ( x ^{2} + 3 ) x}{( x ^{2} + 3 ) ^{2}}

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

\frac{d}{d x} (x^{2} + 3) = 2 x

= 8 \cdot \frac{1 \cdot ( x ^{2} + 3 ) - 2 xx}{( x ^{2} + 3 ) ^{2}}

Simplify

8 \cdot \frac{1 \cdot ( x ^{2} + 3 ) - 2 xx}{( x ^{2} + 3 ) ^{2}} : \frac{8 ( - x ^{2} + 3 )}{( x ^{2} + 3 ) ^{2}}

= \frac{8 ( - x ^{2} + 3 )}{( x ^{2} + 3 ) ^{2}}

\frac{d}{d x} [f (x^{2})] = x^{5}

\frac{\partial}{\partial x} (x^{2} + 2 x y + 3 y^{4})

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

y = ln (\frac{4}{x ^{3}})

\frac{\partial}{\partial x} (x cos (\frac{1}{x ^{2} + y ^{2}}))

What is the derivative of (8x/(x^2+3)) ?
The derivative of (8x/(x^2+3)) is (8(-x^2+3))/((x^2+3)^2)
What is the first derivative of (8x/(x^2+3)) ?
The first derivative of (8x/(x^2+3)) is (8(-x^2+3))/((x^2+3)^2)