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Derivatives of the exponent a^(x)
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I have \[f(x) = 2^{x}-2^{-x}\] could I write it as \[f(x) = 2^{x}- (2^{x})^{-1}\]
yes
If I take the derivative would it be \[f'(x) = 2^{x}\ln2 - [(-1)(\ln2)(2^{x})]\]
2nd term ? i think it isn't correct..
easier thing to do is 2^-x = (2^-1) ^x = (1/2) ^x
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Oh alright then it would ne \[f'(x) = 2^{x}\ln2 - (\frac{1}{2})^{x}\ln(\frac{1}{2})\]
yes, and that comes out to be 2^xln 2 + 2^-x ln 2 = ln 2 (2^x+2^-x)
wait a minute where is your ln 1/2
ln 1/2 = - ln 2
did you use that rule before about -ln2
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yes
so \[f'(0) = 2\ln2\]
correct.
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