The derivative of $2^{x+1}$ in the point $-3$ They ask me to compute the derivative of $2^{x+1}$ in the point $-3$. Since the function is continous in that point, all I have to do is to compute the lateral derivatives. The left one first:
$$\lim_{x \to -3^-} \frac{2^{x+1} - \frac{1}{4}}{x+3}$$
I've tried to solve this limit using this formula:
$$\frac{a^{u(x)} - 1}{u(x)} = \ln{a}$$
But the result seems to be incorrect. How should I solve that limit?

$$\lim_{x \to -3^-} \frac{2^{x+1} - \frac{1}{4}}{x+3} = \lim_{x \to -3^-} \frac{2^{x+1}\frac{2^2}{2^2} - \frac{1}{4}}{x+3} = \lim_{x \to -3^-} \frac{\frac{2^{x+3}}{4} - \frac{1}{4}}{x+3} =$$
$$= \frac{1}{4}\lim_{x \to -3^-} \frac{2^{x+3}- 1}{x+3} = \frac{1}{4}\ln{2}$$
