Question

use the quotient rule to find the derivative of the given function.

\\f(t) = \frac{9t^2 - 2t}{2t + 9}\\

\\f(t) = \\

Explanation:

🆕 New Concept Discovered: Quotient Rule for Differentiation
Differentiating fractions by tracking top and bottom parts

Step 1: Identify the components

We have the function:

$$ f(t) = \frac{9t^2 - 2t}{2t + 9} $$

Let the numerator be \( u(t) \) and the denominator be \( v(t) \):

$$ u(t) = 9t^2 - 2t $$

$$ v(t) = 2t + 9 $$

Now, find their individual derivatives with respect to \( t \):

$$ u'(t) = 18t - 2 $$

$$ v'(t) = 2 $$

Step 2: Apply the Quotient Rule formula

The quotient rule states:

$$ f'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2} $$

Substitute our components into the formula:

$$ f'(t) = \frac{(18t - 2)(2t + 9) - (9t^2 - 2t)(2)}{(2t + 9)^2} $$

Step 3: Expand and simplify the numerator

First, expand the left term in the numerator:

$$ (18t - 2)(2t + 9) = 36t^2 + 162t - 4t - 18 = 36t^2 + 158t - 18 $$

Next, expand the right term in the numerator:

$$ (9t^2 - 2t)(2) = 18t^2 - 4t $$

Subtract the second term from the first:

$$ (36t^2 + 158t - 18) - (18t^2 - 4t) $$

$$ = 36t^2 + 158t - 18 - 18t^2 + 4t $$

Combine like terms:

$$ = 18t^2 + 162t - 18 $$

Step 4: Write the final derivative expression

Place the simplified numerator back over the denominator:

$$ f'(t) = \frac{18t^2 + 162t - 18}{(2t + 9)^2} $$

We can also factor out a common factor of \( 18 \) from the numerator:

$$ f'(t) = \frac{18(t^2 + 9t - 1)}{(2t + 9)^2} $$

Answer:

$$ f'(t) = \frac{18t^2 + 162t - 18}{(2t + 9)^2} $$