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

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

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

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

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

\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.

hows about this one...?

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

hows about this one...?
why don't you try this one so we can see if you understand?

6. Originally Posted by Jhevon
why don't you try this one so we can see if you understand?
ok, I came up with...

(t^(-1/2)-t(^1/2)*t)/(t^2+9)