How do you find the first and second derivative of #y=e^x+e^-x#?

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Answer 1 · 2017-03-02T16:18:46

# dy/dx \ = e^x - e^(-x) #

# (d^2y)/(dx^2) = e^x + e^(-x) #

Using the standard result:

# d/dx e^(ax)=ae^(ax) #

Then if :

# y = e^x+e^(-x) #

We have:

# dy/dx \ = e^x - e^(-x) #

And then:

# (d^2y)/(dx^2) = e^x + e^(-x) #