What is the derivative of ((3x-1/(x^2+3))^2) ?

The derivative of ((3x-1/(x^2+3))^2) is (2(3x-1)(-3x^2+2x+9))/((x^2+3)^3)

What is the first derivative of ((3x-1/(x^2+3))^2) ?

The first derivative of ((3x-1/(x^2+3))^2) is (2(3x-1)(-3x^2+2x+9))/((x^2+3)^3)

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

\frac{2 ( 3 x - 1 ) ( - 3 x ^{2} + 2 x + 9 )}{( x ^{2} + 3 ) ^{3}}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{3 x - 1}{x ^{2} + 3}) = \frac{- 3 x ^{2} + 2 x + 9}{( x ^{2} + 3 ) ^{2}}

= 2 \cdot \frac{3 x - 1}{x ^{2} + 3} \cdot \frac{- 3 x ^{2} + 2 x + 9}{( x ^{2} + 3 ) ^{2}}

Simplify

2 \cdot \frac{3 x - 1}{x ^{2} + 3} \cdot \frac{- 3 x ^{2} + 2 x + 9}{( x ^{2} + 3 ) ^{2}} : \frac{2 ( 3 x - 1 ) ( - 3 x ^{2} + 2 x + 9 )}{( x ^{2} + 3 ) ^{3}}

= \frac{2 ( 3 x - 1 ) ( - 3 x ^{2} + 2 x + 9 )}{( x ^{2} + 3 ) ^{3}}

\int_{1}^{2} \frac{1}{t} d t

\int (x^{2} - 3 x + 3) d x

\int e^{x} + sin (y) d x

x \to 0 lim ((1 + b x)^{\frac{1}{x}})

\int t^{2} + \frac{2 t - 1}{t ^{2} - 2 t + 1} d t

What is the derivative of ((3x-1/(x^2+3))^2) ?
The derivative of ((3x-1/(x^2+3))^2) is (2(3x-1)(-3x^2+2x+9))/((x^2+3)^3)
What is the first derivative of ((3x-1/(x^2+3))^2) ?
The first derivative of ((3x-1/(x^2+3))^2) is (2(3x-1)(-3x^2+2x+9))/((x^2+3)^3)