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What is the derivative of f(x)= ln(e^-x^2)
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\[f(x)=\ln(e^{-x^{2}})\]
You have to use the chain rule. first the derivative of ln(g(x)) = (1/g(x))*g'(x) right?
say g(x) = exp(-x^2)
now what is the derivative of e^h(x)? it is e^h(x) * h'(x)
say h(x) = -x^2
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what is the derivative of h(x)=-x^2? it will be -2x
now plug it all in
derivative of ln(e^(-x^2)) = (1/(e^(-x^2)))*(e^(-x^2))*(-2x)
oh ok so first the derivative of ln(x)= 1/x so that's why 1/e^-x^2 right ...
and then I know that the derivative of e^x is simply e^x unless it's e^(fx) then it is e^(fx) * f'(x)
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I'm getting it now lol ^_^ thanks
that's right!
you have to split all these functions and recognize their derivatives
then you just "chain" it all together
;)
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