So I have the following function:
$f(x)= \sqrt{e^{2x}}$
After applying the chain rule I sit with:
$$\frac{1}{2\sqrt{e^{2x}}}2e^{2x}$$
From there I got:
$$\frac{e^{2x}}{\sqrt{e^{2x}}}$$
While the apparent correct answer is $e^{x}$
If anyone could help explain where I went wrong, or what I am missing I would greatly appreciate it.
$\endgroup$ 24 Answers
$\begingroup$Hint: $\sqrt{e^{2x}} = e^{\left(\dfrac{2x}{2}\right)}$
$\endgroup$ $\begingroup$Hint: $$\sqrt{e^{2x}} = \sqrt{e^{x}e^{x}} = e^x$$
Although you still did get the correct answer of $\frac{e^{2x}}{\sqrt{e^{2x}}}$. With a little algebra you would have found that $$\frac{e^{2x}}{\sqrt{e^{2x}}}=e^x$$
$\endgroup$ $\begingroup$$\sqrt{e^{2x}} = (e^{2x})^{\frac{1}{2}} = e^{x}$
Which has derivative $e^x$
$\endgroup$ 0 $\begingroup$Hint:
$$e^{2x}=e^x \times e^x$$
$\endgroup$
