What is the derivative of (0.6t^2-t+3)/(t^2+3) ?

The derivative of (0.6t^2-t+3)/(t^2+3) is (t^2-2.4t-3)/((t^2+3)^2)

What is the first derivative of (0.6t^2-t+3)/(t^2+3) ?

The first derivative of (0.6t^2-t+3)/(t^2+3) is (t^2-2.4t-3)/((t^2+3)^2)

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derivative of

\frac{0.6 t ^{2} - t + 3}{t ^{2} + 3}

\frac{t ^{2} - 2.4 t - 3}{( t ^{2} + 3 ) ^{2}}

Solution steps

\frac{d}{d t} (\frac{0.6 t ^{2} - t + 3}{t ^{2} + 3})

Apply the Quotient Rule:

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

= \frac{\frac{d}{d t} ( 0.6 t ^{2} - t + 3 ) ( t ^{2} + 3 ) - \frac{d}{d t} ( t ^{2} + 3 ) ( 0.6 t ^{2} - t + 3 )}{( t ^{2} + 3 ) ^{2}}

\frac{d}{d t} (0.6 t^{2} - t + 3) = 1.2 t - 1

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

= \frac{( 1.2 t - 1 ) ( t ^{2} + 3 ) - 2 t ( 0.6 t ^{2} - t + 3 )}{( t ^{2} + 3 ) ^{2}}

Expand

(1.2 t - 1) (t^{2} + 3) - 2 t (0.6 t^{2} - t + 3) : t^{2} - 2.4 t - 3

= \frac{t ^{2} - 2.4 t - 3}{( t ^{2} + 3 ) ^{2}}

\frac{d}{d x} (8 x - 1)

10 x y y^{^{'}} = 1 - y^{2}

x \to \infty lim (\frac{x ^{3} + 9 x}{6 x ^{3} - x ^{2} + 3})

\frac{d}{d x} (6 x - 1)

\int_{- 2}^{2} x^{4} - 8 x^{2} + 16 d x

What is the derivative of (0.6t^2-t+3)/(t^2+3) ?
The derivative of (0.6t^2-t+3)/(t^2+3) is (t^2-2.4t-3)/((t^2+3)^2)
What is the first derivative of (0.6t^2-t+3)/(t^2+3) ?
The first derivative of (0.6t^2-t+3)/(t^2+3) is (t^2-2.4t-3)/((t^2+3)^2)