Originally Posted by

**Craka** Question asks to find k in the form of a limit of

$\displaystyle

\begin{array}{l}

y = 4^x \\

\frac{{dy}}{{dx}} = k.4^x \\

\end{array}

$

This is my attempt

$\displaystyle

\begin{array}{l}

\frac{{dy}}{{dx}} = \mathop {\lim }\limits_{h \to 0} \frac{{4^{a + h} {\color{red}-} 4^a }}{h} \\

k.4^a = \frac{{4^{a + h} {\color{red}-} 4^a }}{h} \\

k = \frac{1}{{4^a }}\frac{{4^{a + h} {\color{red}-} 4^a }}{h} \\

\end{array}

$

The answer is given as

$\displaystyle

\mathop {\lim }\limits_{h \to 0} \frac{{4^h {\color{red}-} 1}}{h}

$

I'm not sure how to get to this.