What is the derivative of y=(4x^2+6x)/(3x^3+2) ?

The derivative of y=(4x^2+6x)/(3x^3+2) is (-12x^4-36x^3+16x+12)/((3x^3+2)^2)

What is the first derivative of y=(4x^2+6x)/(3x^3+2) ?

The first derivative of y=(4x^2+6x)/(3x^3+2) is (-12x^4-36x^3+16x+12)/((3x^3+2)^2)

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derivative of

y = \frac{4 x ^{2} + 6 x}{3 x ^{3} + 2}

\frac{- 12 x ^{4} - 36 x ^{3} + 16 x + 12}{( 3 x ^{3} + 2 ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

\frac{d}{d x} (4 x^{2} + 6 x) = 8 x + 6

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

= \frac{( 8 x + 6 ) ( 3 x ^{3} + 2 ) - 9 x ^{2} ( 4 x ^{2} + 6 x )}{( 3 x ^{3} + 2 ) ^{2}}

Expand

(8 x + 6) (3 x^{3} + 2) - 9 x^{2} (4 x^{2} + 6 x) : - 12 x^{4} - 36 x^{3} + 16 x + 12

= \frac{- 12 x ^{4} - 36 x ^{3} + 16 x + 12}{( 3 x ^{3} + 2 ) ^{2}}

x \to \infty lim ((\frac{( e ^{x} - e ^{- x} )}{( e ^{x} + e ^{- x} )}) + 4)

\int (5 cos (x) - 4, x) d x

x \to \infty lim (\frac{2 - x}{x + 1} + \frac{2 ^{- x} \cdot x}{x + 2})

y^{^{'}} = 2 y (1 - \frac{y}{6200})

\int_{0}^{1} \frac{18 x}{x ^{2} + 1} d x

What is the derivative of y=(4x^2+6x)/(3x^3+2) ?
The derivative of y=(4x^2+6x)/(3x^3+2) is (-12x^4-36x^3+16x+12)/((3x^3+2)^2)
What is the first derivative of y=(4x^2+6x)/(3x^3+2) ?
The first derivative of y=(4x^2+6x)/(3x^3+2) is (-12x^4-36x^3+16x+12)/((3x^3+2)^2)