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

The derivative of y=(4-3x)/(3x^2+x) is (9x^2-24x-4)/((3x^2+x)^2)

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

The first derivative of y=(4-3x)/(3x^2+x) is (9x^2-24x-4)/((3x^2+x)^2)

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derivative of

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

\frac{9 x ^{2} - 24 x - 4}{( 3 x ^{2} + x ) ^{2}}

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Simplify

\frac{( - 3 ) ( 3 x ^{2} + x ) - ( 6 x + 1 ) ( 4 - 3 x )}{( 3 x ^{2} + x ) ^{2}} : \frac{9 x ^{2} - 24 x - 4}{( 3 x ^{2} + x ) ^{2}}

= \frac{9 x ^{2} - 24 x - 4}{( 3 x ^{2} + x ) ^{2}}

630 - 4.9 t^{2}

x \to \frac{π}{2} lim (8 sec (x))

y^{^{'}} = y (x y^{4} + 3)

y^{^{'}} - \frac{2 y}{x} = 1

9^{^{'}}

What is the derivative of y=(4-3x)/(3x^2+x) ?
The derivative of y=(4-3x)/(3x^2+x) is (9x^2-24x-4)/((3x^2+x)^2)
What is the first derivative of y=(4-3x)/(3x^2+x) ?
The first derivative of y=(4-3x)/(3x^2+x) is (9x^2-24x-4)/((3x^2+x)^2)