1. ## exponential function derivative

Can anyone help me find the derivative of y=2^|x|?

2. Originally Posted by nejikun
Can anyone help me find the derivative of y=2^|x|?
If $\displaystyle f(x) = a^x$ then $\displaystyle f'(x) = a^xln(a)$

Spoiler:
Take the log of both sides:

$\displaystyle ln(y) = ln(2^{|x|}) = |x|ln(2)$

Differentiate both sides:

$\displaystyle \frac{1}{y} \cdot \frac{dy}{dx} = ln(2)$

Isolate dy/dx:

$\displaystyle \frac{dy}{dx} = yln(2) = 2^{|x|}ln(2)$

3. Originally Posted by nejikun
Can anyone help me find the derivative of y=2^|x|?
Hi nejikun.

$\displaystyle y\ =\ 2^{|x|}\ =\ e^{(\ln2)|x|}\ =\ e^u$ where $\displaystyle u=(\ln2)|x|$

So, using the chain rule, $\displaystyle \frac{dy}{dx}\ =\ \frac{dy}{du}\,\frac{du}{dx}\ =\ e^u\cdot(\ln2)\frac{|x|}x\ =\ \frac{(\ln2)|x|}x2^{|x|}.$

Note that $\displaystyle \frac{d\left(|x|\right)}{dx}=\frac{|x|}x.$

4. Originally Posted by e^(i*pi)
$\displaystyle ln(y) = ln(2^{|x|}) = |x|ln(2)$

Differentiate both sides:

$\displaystyle \frac{1}{y} \cdot \frac{dy}{dx} = ln(2)$
The derivative of $\displaystyle |x|$ is 1 if $\displaystyle x>0$ and $\displaystyle -1$ if $\displaystyle x<0;$ i.e. $\displaystyle \frac{d(|x|)}{dx}=\frac{|x|}x.$