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#