How do you find the derivative of #y=e^(e^(3x^2))#?

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Answer 1 · 2016-10-24T12:13:57

#y'=6xe^(e^(3x^2)+3x^2)#

The rule for differentiating e functions is,

#y=e^f(x)#

#y'=f'(x)e^f(x)#

so in the case of your question we have

#y=e^f(x)# where #f(x) = e^g(x)#

so we will differentiate to find each part we need,
#y=e^f(x)#

#f(x) = e^g(x)#

#f'(x)=g'(x)e^g(x)#

So now what is y, f(x) and g(x).

#y=e^(e^(3x^2))#

#f(x)=e^(3x^2)#

#g(x)=e^(3x^2)#

#f(x)=e^(3x^2)#

#f'(x)= 6x*e^(3x^2)#

so subbing into the original formula,

#y'=f'(x)e^f(x)#

#y'=(6x*e^(3x^2))(e^(e^(3x^2))) #

#y'=6x*e^(e^(3x^2))*e^(3x^2)#

#y'=6xe^(e^(3x^2)+3x^2)#