What is the derivative of (t^3+3t)/(t^2-4t+3) ?

The derivative of (t^3+3t)/(t^2-4t+3) is (t^4-8t^3+6t^2+9)/((t^2-4t+3)^2)

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

The first derivative of (t^3+3t)/(t^2-4t+3) is (t^4-8t^3+6t^2+9)/((t^2-4t+3)^2)

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derivative of

\frac{t ^{3} + 3 t}{t ^{2} - 4 t + 3}

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

Solution steps

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

Apply the Quotient Rule:

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

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

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

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

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

Expand

(3 t^{2} + 3) (t^{2} - 4 t + 3) - (2 t - 4) (t^{3} + 3 t) : t^{4} - 8 t^{3} + 6 t^{2} + 9

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

\int \frac{7}{x ( x - 7 ) ^{2}} d x

\frac{d}{d x} ((x - 100)^{3} + (x - 100)^{2} + 500)

\int 1 - y^{2} d y

\frac{d}{d x} ((x^{3} + 7)^{\frac{2}{3}})

g (x) = 8 - x

What is the derivative of (t^3+3t)/(t^2-4t+3) ?
The derivative of (t^3+3t)/(t^2-4t+3) is (t^4-8t^3+6t^2+9)/((t^2-4t+3)^2)
What is the first derivative of (t^3+3t)/(t^2-4t+3) ?
The first derivative of (t^3+3t)/(t^2-4t+3) is (t^4-8t^3+6t^2+9)/((t^2-4t+3)^2)