Let \( f(x)=4^{x} \) and \( g(x)=\left(\frac{1}{3}\right)^{x} \), and evaluate the following: \( \frac{f(x)-f(3)}{x-3}=\square \)

Cargar el desarrollo paso a paso de la derivada de \( f(x)=\operatorname{Sen}^{2}\left(2 x^{3}\right) \) es

\( \int x \operatorname { arcsen } x d x \)

21. \( -\int(x-1)^{3} d x \) 22.- \( \int \frac{x}{x^{2}-1} d x \) 23.- \( \int \frac{4 x-3}{2 x^{2}-3 x-4} d x \) 24. \( -\int x^{2} \cdot e^{x^{3}} d x \) 25.- \( \int e^{2 x} d x \) 26. \( -\int 5^{3 x} d x \) 27. \( -\int \frac{\operatorname{sen} x}{5+e^{\cos x}} d x \) 28. \( -\int \sin 2 x d x \) 29.- \( \int x \cos x^{2} d x \) 30.- \( \int 4 \sin 5 x d x \)

(1) \( \lim _{(x, y) \rightarrow(0,1)} \frac{x-x y+3}{x^{2} y+5 x y-y^{3}}= \)

8. (15 pts) Let \( f(x, y)=x^{3}+3 y^{2} x-36 y-39 x+3 \). This function has four critical points at \( (-2,-3),(-3,-2),(2,3) \), and \( (3,2) \). (You don't need to show that they are critical points) Use the Second Derivative Test to find the local maximum and minimum values and saddle points of the function.