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

Answer 1

Emilia Aguilar

# 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) #

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Answer 2

Isaiah Allen

To find the derivatives of ( y = e^x + e^{-x} ):

First derivative: [ \frac{dy}{dx} = e^x - e^{-x} ]

Second derivative: [ \frac{d^2y}{dx^2} = e^x + e^{-x} ]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.