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

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

We have:

Differentiating directly:

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

and so:

By signing up, you agree to our Terms of Service and Privacy Policy

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} ).

By signing up, you agree to our Terms of Service and Privacy Policy

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.

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7