How do you find the derivative of $2^x$ ?

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Answer 1 · 2018-04-06T18:39:01

$d/dx2^x=2^xln2$

In general, the derivative of an exponential with some constant base is

$d/dxa^x=a^xlna$ .

A proof of this will be shown.

So, $d/dx2^x=2^xln2$

Proof:

Rewrite $a^x$ as $e^ln(a^x)=e^(xlna)$ .

Now,

$d/dxa^x=d/dxe^(xlna)=e^(xlna)*d/dx(xlna),$ as per the Chain Rule.

We then have

$d/dxa^x=e^(xlna)ln(a),$ and, recalling that $e^(xlna)=a^x,$ we finally have

$d/dxa^x=a^xlna$ .

Answer 2 · 2018-04-06T19:20:51

Let $y=2^x$

Taking log on both sides,

$logy=log2^x$

$=>logy=xlog2$

Applying derivative,

$=>1/y dy/dx=log2$

$=>dy/dx=2^xlog2$

How do you find the derivative of 2^x2^x?