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

Answer 1

Evelyn Agard

#= (e^x+e^-x) / 2#

#d/dx ((e^x-e^-x) / 2)#

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

#= e^x/2 - (-e^-x) / 2#

#= (e^x+e^-x) / 2#

or if you like.....

#d/dx ((e^x-e^-x) / 2)#

#= d/dx sinh x#

#= cosh x#

#= (e^x+e^-x) / 2#

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Answer 2

Charles Anctil

To differentiate the function ( f(x) = \frac{e^x - e^{-x}}{2} ), you can use the quotient rule. The quotient rule states that for functions ( u(x) ) and ( v(x) ), the derivative of ( \frac{u(x)}{v(x)} ) with respect to ( x ) is given by:

[ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2} ]

Now, let's apply this rule to the function ( f(x) = \frac{e^x - e^{-x}}{2} ):

[ u(x) = e^x - e^{-x} ] [ v(x) = 2 ]

[ u'(x) = e^x + e^{-x} ] [ v'(x) = 0 ]

Now substitute these values into the quotient rule formula:

[ f'(x) = \frac{(e^x + e^{-x})(2) - (e^x - e^{-x})(0)}{(2)^2} ]

[ f'(x) = \frac{2e^x + 2e^{-x}}{4} ]

[ f'(x) = \frac{e^x + e^{-x}}{2} ]

So, the derivative of ( f(x) = \frac{e^x - e^{-x}}{2} ) with respect to ( x ) is ( f'(x) = \frac{e^x + e^{-x}}{2} ).

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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.