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

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

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Answer 1 · 2018-04-06T18:39:01

$\frac{d}{d x} 2^{x} = 2^{x} ln 2$ $d/dx2^x=2^xln2$

Explanation:

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

$\frac{d}{d x} a^{x} = a^{x} ln a$ $d/dxa^x=a^xlna$ .

A proof of this will be shown.

So, $\frac{d}{d x} 2^{x} = 2^{x} ln 2$ $d/dx2^x=2^xln2$

Proof:

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

Now,

$\frac{d}{d x} a^{x} = \frac{d}{d x} e^{x ln a} = e^{x ln a} \cdot \frac{d}{d x} (x ln a),$ $d/dxa^x=d/dxe^(xlna)=e^(xlna)*d/dx(xlna),$ as per the Chain Rule.

We then have

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

$\frac{d}{d x} a^{x} = a^{x} ln a$ $d/dxa^x=a^xlna$ .

Answer 2 · 2018-04-06T19:20:51

Let $y = 2^{x}$ $y=2^x$

Taking log on both sides,

$log y = {log 2}^{x}$ $logy=log2^x$

$\Rightarrow log y = x log 2$ $=>logy=xlog2$

Applying derivative,

$\Rightarrow \frac{1}{y} \frac{d y}{d x} = log 2$ $=>1/y dy/dx=log2$

$\Rightarrow \frac{d y}{d x} = 2^{x} log 2$ $=>dy/dx=2^xlog2$

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

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