What is the derivative of (4x/(x^2+3)) ?

The derivative of (4x/(x^2+3)) is (4(-x^2+3))/((x^2+3)^2)

What is the first derivative of (4x/(x^2+3)) ?

The first derivative of (4x/(x^2+3)) is (4(-x^2+3))/((x^2+3)^2)

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

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

Solution steps

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

Take the constant out:

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

= 4 \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}}

= 4 \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

= 4 \cdot \frac{1 \cdot ( x ^{2} + 3 ) - 2 xx}{( x ^{2} + 3 ) ^{2}}

Simplify

4 \cdot \frac{1 \cdot ( x ^{2} + 3 ) - 2 xx}{( x ^{2} + 3 ) ^{2}} : \frac{4 ( - x ^{2} + 3 )}{( x ^{2} + 3 ) ^{2}}

= \frac{4 ( - x ^{2} + 3 )}{( x ^{2} + 3 ) ^{2}}

\frac{d y}{d x} = \frac{cos ( 7 x )}{e ^{7 y}}

\int cos (x) sin (x) - x y^{2} d x

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

f (x) = x sin (x)

\int \frac{24}{24 + e ^{x}} d x

What is the derivative of (4x/(x^2+3)) ?
The derivative of (4x/(x^2+3)) is (4(-x^2+3))/((x^2+3)^2)
What is the first derivative of (4x/(x^2+3)) ?
The first derivative of (4x/(x^2+3)) is (4(-x^2+3))/((x^2+3)^2)