Find the derivative of $$ y=\frac{t^{2}}{3 t^{2}-7 t+10} $$ $$ y^{\prime}=\square $$

Question

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To find the derivative of the function

$$
y = \frac{t^2}{3t^2 - 7t + 10},
$$

we will use the quotient rule. The quotient rule states that if you have a function in the form $$ y = \frac{u}{v} $$, then the derivative $$ y' $$ is given by:

$$
y' = \frac{u'v - uv'}{v^2},
$$

where $$ u = t^2 $$ and $$ v = 3t^2 - 7t + 10 $$.

First, we need to find $$ u' $$ and $$ v' $$:

1. **Calculate $$ u' $$**:
   $$
   u = t^2 \implies u' = 2t.
   $$

2. **Calculate $$ v' $$**:
   $$
   v = 3t^2 - 7t + 10 \implies v' = 6t - 7.
   $$

Now we can apply the quotient rule:

$$
y' = \frac{(2t)(3t^2 - 7t + 10) - (t^2)(6t - 7)}{(3t^2 - 7t + 10)^2}.
$$

Next, we will simplify the numerator:

1. Expand $$ (2t)(3t^2 - 7t + 10) $$:
   $$
   2t(3t^2) - 2t(7t) + 2t(10) = 6t^3 - 14t^2 + 20t.
   $$

2. Expand $$ (t^2)(6t - 7) $$:
   $$
   t^2(6t) - t^2(7) = 6t^3 - 7t^2.
   $$

Now substitute these expansions back into the expression for $$ y' $$:

$$
y' = \frac{(6t^3 - 14t^2 + 20t) - (6t^3 - 7t^2)}{(3t^2 - 7t + 10)^2}.
$$

Now simplify the numerator:

$$
y' = \frac{6t^3 - 14t^2 + 20t - 6t^3 + 7t^2}{(3t^2 - 7t + 10)^2}.
$$

Combine like terms:

$$
y' = \frac{-14t^2 + 7t^2 + 20t}{(3t^2 - 7t + 10)^2} = \frac{-7t^2 + 20t}{(3t^2 - 7t + 10)^2}.
$$

Thus, the derivative is:

$$
y' = \frac{-7t^2 + 20t}{(3t^2 - 7t + 10)^2}.
$$

So, the final answer is:

$$
y' = \frac{-7t^2 + 20t}{(3t^2 - 7t + 10)^2}.
$$
$$ y' = \frac{-7t^2 + 20t}{(3t^2 - 7t + 10)^2} $$

Find the derivative of

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