How do you find the derivative of #y=(e^x+e^-x)/4#?

Answer 1

# dy/dx = (e^x-e^(-x))/4 = 1/2sinhx #

We have:

# y = (e^x+e^(-x))/4 #

Differentiating directly:

# dy/dx = (e^x-e^(-x))/4 #

Also, If you are familiar with the hyperbolic functions then we can proceed as follows:

# y = (e^x+e^(-x))/4 #
# \ \ = 1/2 * (e^x+e^(-x))/2 #
# \ \ = 1/2coshx #

and so:

# dy/dx = 1/2sinhx #
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Answer 2

To find the derivative of ( y = \frac{e^x + e^{-x}}{4} ), you can use the quotient rule. The quotient rule states that if ( y = \frac{u}{v} ), then ( y' = \frac{u'v - uv'}{v^2} ). In this case:

Let ( u = e^x + e^{-x} ) and ( v = 4 ).

Then, ( u' = e^x - e^{-x} ) and ( v' = 0 ) (since the derivative of a constant is zero).

Now apply the quotient rule:

[ y' = \frac{(e^x - e^{-x}) \cdot 4 - (e^x + e^{-x}) \cdot 0}{4^2} ]

[ y' = \frac{4(e^x - e^{-x})}{16} ]

[ y' = \frac{e^x - e^{-x}}{4} ]

Therefore, the derivative of ( y = \frac{e^x + e^{-x}}{4} ) is ( y' = \frac{e^x - e^{-x}}{4} ).

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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.

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.

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.

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.

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