A particle is moving with acceleration \( a(t)=18 t+12 \). its position at time \( t=0 \) is \( s(0)=12 \) and its velocity at time \( t=0 \) is \( v(0)=3 \). What is its position at time \( t=7 \) ?

The equation implicitly defines \( u \) as a function of \( x \). \( \quad-\frac{x^{2}}{u^{2}}-\cos (6 \mathrm{u})=-4 \) Find \( \frac{d u}{d x} \).

7. Calcular \( \frac{d}{d x} \int_{1}^{x} \frac{1}{t} d t \) \( \begin{array}{ll}\ln t & \text { b) } \frac{1}{t} \\ \text { c) } \frac{1}{x} & \text { d) } \frac{\ln x}{x}\end{array} \)

f) \( \lim _{x \rightarrow 0} \frac{\tan x+x^{2}}{(\sin x)^{3}+2 x} \)

\( \lim _{x \rightarrow 2} f(x) \) Siendo f la siguiente función definida por partes: \( f(x)=\left\{\begin{array}{l}x^{2}, \text { si } x<2 \\ \frac{1}{x}, \\ \text { si } x \geq 2\end{array}\right. \)

32. \( y=\operatorname{sen} x-x \cos x+x^{2}+4 x+3 \) Solucion: \( x \operatorname{sen} x+2 x+4 \)