# Thread: derivatives using the quotient rule

1. ## derivatives using the quotient rule

Find the derivative of the function below.

a little help por favor

2. Hello, mer1988!

Differentiate: .$\displaystyle z(t)\:=\:\frac{t^2+5t+2}{t+3}$
Quotient Rule . Given: $\displaystyle y \:=\:\frac{f}{g}$
. . . . . . . . . . . .then: .$\displaystyle y' \:=\:\frac{gf' - fg'}{g^2}$

We have: .$\displaystyle z(t) \;=\;\frac{\overbrace{t^2+5t+2}^{f}}{\underbrace{\ ,t+3\,}_{g}}$

Then: .$\displaystyle z'(t) \;=\;\frac{\overbrace{(t+3)}^{g}\overbrace{(2t+5)} ^{f'} - \overbrace{(t^2+5t+2)}^f\overbrace{(1)}^{g'}}{\und erbrace{(t+3)^2}_{g^2}} \;= \;\frac{2t^2+5t+6t+15 - t^2-5t-2}{(t+3)^2}$

Therefore: .$\displaystyle z'(t)\;=\;\frac{t^2+6t+13}{(t+3)^2}$

3. Originally Posted by mer1988
Si no quieres usar la regla del cuociente, también podemos hacer lo siguiente

\displaystyle \begin{aligned} z(t) &= \frac{{t^2 + 5t + 2}} {{t + 3}}\\ &= \frac{{t(t + 3) + 2(t + 3) - 4}} {{t + 3}}\\ &= t + 2 - 4(t + 3)^{ - 1}\\ z'(t) &= 1 + \frac{4} {{(t + 3)^2 }}\\ &= \frac{{t^2 + 6t + 13}} {{(t + 3)^2 }} \end{aligned}

Saludos

4. Find the derivative of the function below.

5. Originally Posted by mer1988
Find the derivative of the function below.