How do you differentiate #f(x)=2^(-x^2)#?

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Answer 1 · 2017-08-17T00:09:55

Please see below.

For #a > 0#, #d/dx(a^x) = a^x lna#

For exponent other then #x#, we use the chain rule, so that

#d/dx(a^u) = a^u ln(a) (du)/dx#

So for this question, we have

#f'(x) = 2^(-x^2) ln(2) * (-2x)#

# = -2x2^(-x^2)ln(2)#

Or #-x2^(-x^2+1)ln(2)#