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

The derivative of ((2x^2+1/(3x^3+1))^2) is (2(2x^2+1)(-6x^4-9x^2+4x))/((3x^3+1)^3)

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

The first derivative of ((2x^2+1/(3x^3+1))^2) is (2(2x^2+1)(-6x^4-9x^2+4x))/((3x^3+1)^3)

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

\frac{2 ( 2 x ^{2} + 1 ) ( - 6 x ^{4} - 9 x ^{2} + 4 x )}{( 3 x ^{3} + 1 ) ^{3}}

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (\frac{2 x ^{2} + 1}{3 x ^{3} + 1}) = \frac{- 6 x ^{4} - 9 x ^{2} + 4 x}{( 3 x ^{3} + 1 ) ^{2}}

= 2 \cdot \frac{2 x ^{2} + 1}{3 x ^{3} + 1} \cdot \frac{- 6 x ^{4} - 9 x ^{2} + 4 x}{( 3 x ^{3} + 1 ) ^{2}}

Simplify

2 \cdot \frac{2 x ^{2} + 1}{3 x ^{3} + 1} \cdot \frac{- 6 x ^{4} - 9 x ^{2} + 4 x}{( 3 x ^{3} + 1 ) ^{2}} : \frac{2 ( 2 x ^{2} + 1 ) ( - 6 x ^{4} - 9 x ^{2} + 4 x )}{( 3 x ^{3} + 1 ) ^{3}}

= \frac{2 ( 2 x ^{2} + 1 ) ( - 6 x ^{4} - 9 x ^{2} + 4 x )}{( 3 x ^{3} + 1 ) ^{3}}

sin (t) = y^{^{'}} + y

n = 1 \sum \infty \frac{- 1}{n + 1}

\int (1 - 6 x) e^{3 x - 9 x^{2}} d x

n \to \infty lim (7^{n})

What is the derivative of ((2x^2+1/(3x^3+1))^2) ?
The derivative of ((2x^2+1/(3x^3+1))^2) is (2(2x^2+1)(-6x^4-9x^2+4x))/((3x^3+1)^3)
What is the first derivative of ((2x^2+1/(3x^3+1))^2) ?
The first derivative of ((2x^2+1/(3x^3+1))^2) is (2(2x^2+1)(-6x^4-9x^2+4x))/((3x^3+1)^3)