What is the derivative of $2^{2 x}$ $2^(2x)$ ?

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Answer 1 · 2016-06-05T11:58:23

$= 2^{2 x + 1} ln (2)$ $=2^{2x+1}ln (2)$

$\frac{d}{d x} (2^{2 x})$ $frac{d}{dx}(2^{2x})$
Applying exponent rule, $a^{b} = e^{b ln (a)}$ $a^b=e^{bln (a)}$

$2^{2 x} = e^{2 x ln (2)}$ $2^{2x}=e^{2xln (2)}$
$= \frac{d}{d x} (e^{2 x ln (2)})$ $=frac{d}{dx}(e^{2x\ln (2)})$

Applying chain rule,
$\frac{d f (u)}{d x} = \frac{d f}{d u} \cdot \frac{d u}{d x}$ $frac{df(u)}{dx}=frac{df}{du}cdot frac{du}{dx}$

Let, $2 x ln (2) = u$ $2xln (2)=u$
$= \frac{d}{d u} (e^{u}) \frac{d}{d x} (2 x ln (2))$ $=frac{d}{du}(e^u)frac{d}{dx}(2xln (2))$

We know,
$\frac{d}{d u} (e^{u}) = e^{u}$ $frac{d}{du}(e^u)=e^u$
and,
$\frac{d}{d x} (2 x ln (2)) = 2 ln (2)$ $frac{d}{dx}(2xln (2))=2ln (2)$

Also,
$= e^{u} 2 ln (2)$ $=e^u2ln (2)$

Substituting back, $u = 2 x ln (2)$ $u=2xln (2)$

Simplifying it,
$= 2^{2 x + 1} ln (2)$ $=2^{2x+1}ln (2)$

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