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

The derivative of f(x)=(2x^3+4x)/(x^2+3) is (2x^4+14x^2+12)/((x^2+3)^2)

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

The first derivative of f(x)=(2x^3+4x)/(x^2+3) is (2x^4+14x^2+12)/((x^2+3)^2)

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derivative of

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

\frac{2 x ^{4} + 14 x ^{2} + 12}{( x ^{2} + 3 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (2 x^{3} + 4 x) = 6 x^{2} + 4

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

= \frac{( 6 x ^{2} + 4 ) ( x ^{2} + 3 ) - 2 x ( 2 x ^{3} + 4 x )}{( x ^{2} + 3 ) ^{2}}

Expand

(6 x^{2} + 4) (x^{2} + 3) - 2 x (2 x^{3} + 4 x) : 2 x^{4} + 14 x^{2} + 12

= \frac{2 x ^{4} + 14 x ^{2} + 12}{( x ^{2} + 3 ) ^{2}}

\int_{- 1}^{3} (1 + 3 x) d x

\int \frac{1}{x 10 x + 49} d x

\frac{d}{d x} (2 x^{2} + 3 x + 1)

\int \frac{5 x}{16 - 4 x ^{2}} d x

\int_{12}^{30} \frac{1}{30} d x

What is the derivative of f(x)=(2x^3+4x)/(x^2+3) ?
The derivative of f(x)=(2x^3+4x)/(x^2+3) is (2x^4+14x^2+12)/((x^2+3)^2)
What is the first derivative of f(x)=(2x^3+4x)/(x^2+3) ?
The first derivative of f(x)=(2x^3+4x)/(x^2+3) is (2x^4+14x^2+12)/((x^2+3)^2)