$$f' = \dfrac{e^{-5x} * cos(3x)-e^{-5x}*[cos(3x)]'}{cos^2(3x)}$$

$${ df(x)} over { dx } = { { e^{(-5x)'} } * cos( 3x) - e^{-5x} * [ cos(3x) ]' } over { cos^2(3x) } = { e^{-5x}*(-5 * cos(3x) 3sen(-3x))} over { cos^2(3x) } }$$


## 2.
$$
\large
y=x^3 \cdot(\sqrt{x} + x^2)
$$
$$
y=x^3 \cdot(x^{\frac{1}{2}} + x^2)
$$
$$
y=x^{\frac{3}{2}} + x^5
$$
$$
y=x^{\frac{7}{2}} + x^5
$$
$$
y'=\frac{7}{2} \cdot x^{\frac{5}{2}} + 5\cdot x^4
$$
$$
y'=\small \dfrac{7}{2}. x^2\sqrt{x}+5x^4
$$
# 3. 
$f(x) = \sqrt{2(x^2 - 8)}+x$ se $x<-4$
$f(x) = 2^{x+4}$ se $x\ge-4$

# 4.
**Coeficiente angular:**
$3y = 2 - x$
$y = \frac{2}{3} - \frac{x}{3}$

$r_y = b-\frac{x}{3}$

Para tingir (com tinta :) a derivada deve ser $-\frac{1}{3}$
$f'(x) = 8x-3$
$f'(x) = 8x-3$

$8x-3 = -\frac{1}{3}$
$8x = 3-\frac{1}{3}$
$8x = \frac{8}{3}$
$x = \frac{1}{3}$

Encontrar o ponto de tangência
$f(\frac{1}{3}) = 4\cdot(\frac{1}{3})^2 - 1$
$f(\frac{1}{3}) = \frac{4}{9} - 1$
$f(\frac{1}{3}) = -\frac{5}{9}$

Equação da reta:
$$y = mx+ b$$
$