# Derivative Help: sinhx/coshx

• ldbaseball16
In summary: If I wrote 3 + 1/4, most anyone would interpret this as 3 1/4, not as 4/4 = 1."You've also written sinhx and coshx on one line. This is not allowed. To write these functions properly, you need to separate them out by Putting a comma after sinhx and another after coshx. "To write these functions properly, you need to separate them out by Putting a comma after sinhx and another after coshx."After you've written the functions and separated them out, you can try to find the derivative. "After you've written the functions and separated them out, you can try to find

derivative help!

## Homework Statement

find f'(x) sinhx/coshx

## Homework Equations

this is what i got but I am unsure if its wrong could anyone help me pleasee

## The Attempt at a Solution

coshx(sinhx)-(coshx)coshx/coshx... is this right if not help please

How are you getting that? Write down the quotient rule for the derivative of f(x)/g(x) and check the parts.

ok i took a look at the quotient rule and was completely wrong how does this look...

coshx(coshx)-sinhx(sinhx)/(coshx)^2..... (Coshx)^2-(Sinhx)^2/(Coshx)^2...the (coshx)^2 cancel out and you are left with -(sinhx)^2?

Still not right. Cosh(x)^2 does not simply cancel out and vanish. If I asked you what 5/5 was equivalent to would you say zero?

To get the correct answer use the identity cosh(x)^2 - sinh(x)^2 = 1 or the identity 1 - tanh(x)^2 = sech(x)^2

ooooooo the identity oo ok so its 1/(coshx)^2?

Can you simplify that further?

no i get zero if i do the quotient rule again...?

No, you don't need to apply the quotient rule again nor should you; moreover, the derivative of 1/cosh(x)^2 is not zero.

Maybe you haven't been taught this but sech(x) = 1/cosh(x). Now simplify it.

oooo yessss i found it in my notebook in the hyperbolic functions would it be sech^2x??

Yes, d(tanh(x))/dx = sech(x)^2.

ldbaseball16 said:

## Homework Statement

find f'(x) sinhx/coshx
You've come to this Web site, so I suppose that means you would like some help. You've taken the first step, which is to show us what you're tried to do. Good.

One thing you can do to help your cause is to provide meaningful information.
"find f'(x) sinhx/coshx"
Given just this information, I wouldn't know what to do. You're asking us to find the derivative of some function, but you haven't explicitly told us what the function is. Of course, we might infer that f(x) = sinhx/coshx, and you want to find f'(x).
ldbaseball16 said:

## Homework Equations

this is what i got but I am unsure if its wrong could anyone help me pleasee

## The Attempt at a Solution

coshx(sinhx)-(coshx)coshx/coshx... is this right if not help please

It's harder to write mathematical expressions on a single line than on paper, where you can write fractions more easily. When you have to write a complicated expression with fractions on a single line, make sure that you put a pair of parentheses around the entire numerator and another around the entire denominator. For example, if I write 3 + 1/4, most anyone would interpret this as 3 1/4, not as 4/4 = 1.

## 1. What is the derivative of sinh(x)?

The derivative of sinh(x) is cosh(x).

## 2. What is the derivative of cosh(x)?

The derivative of cosh(x) is sinh(x).

## 3. How do you find the derivative of sinh(x)/cosh(x)?

To find the derivative of sinh(x)/cosh(x), use the quotient rule and simplify the resulting expression.

## 4. Is there a shortcut for finding the derivative of sinh(x)/cosh(x)?

Yes, there is a shortcut called the hyperbolic quotient rule, which states that the derivative of sinh(x)/cosh(x) is equal to 1 - tanh(x)^2.

## 5. Can I use the chain rule to find the derivative of sinh(x)/cosh(x)?

Yes, you can also use the chain rule to find the derivative of sinh(x)/cosh(x). The resulting expression will be the same as using the hyperbolic quotient rule.