- #1

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Need help in the steps to take to find the derivative of

y=cosh(x^2 + loge(x))

and e^y +ytanh(x) = x

I have never seen them before, therefor not sure which rule to use, im thinking the second needs partial derivatives

Thanks!!

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- Thread starter Rizzamabob
- Start date

- #1

- 21

- 0

Need help in the steps to take to find the derivative of

y=cosh(x^2 + loge(x))

and e^y +ytanh(x) = x

I have never seen them before, therefor not sure which rule to use, im thinking the second needs partial derivatives

Thanks!!

- #2

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- 0

In my calculus book the rules for differentiating and integrating hyperbolic functions are buried in a chapter, rather than listed on the front inside cover.

I bet your book is set up in the same way.

- #3

Tom Mattson

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Rizzamabob said:y=cosh(x^2 + loge(x))

Is the part in red supposed to be [itex]\log_e(x)[/itex] (as in [itex]\ln(x)[/itex]), or is that supposed to be [itex]\log\left(e^x\right)[/itex]?

- #4

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Left my book at uni, thats why i need help now :(

- #5

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[itex]\log_e(x)[/itex]Tom Mattson said:And one other thing...

Is the part in red supposed to be [itex]\log_e(x)[/itex] (as in [itex]\ln(x)[/itex]), or is that supposed to be [itex]\log\left(e^x\right)[/itex]?

Thanks

- #6

Tom Mattson

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Rizzamabob said:Left my book at uni, thats why i need help now :(

Google, buddy, Google.

We're happy to help you with your questions, but as per the notice at the top of this Forum, you have to make

Wikipedia

Math World

You will find definitions to all things mathematical at those sites. So, please search for "hyperbolic functions", compose your thoughts on the problem, and let us know where you're stuck. You'll be better off for it!

- #7

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Thanks, atm im working this way

y = (x^2 (e^x + e^-x))/2 + ([itex]\log_e(x)[/itex](e^x + e^-x))/2

===>

y = ((e^x + e^-x) (x^2 + [itex]\log_e(x)[/itex]))/2

:surprised

Then try to find dy/dx i have a feeling it will be YUCK :surprised

y = (x^2 (e^x + e^-x))/2 + ([itex]\log_e(x)[/itex](e^x + e^-x))/2

===>

y = ((e^x + e^-x) (x^2 + [itex]\log_e(x)[/itex]))/2

:surprised

Then try to find dy/dx i have a feeling it will be YUCK :surprised

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- #8

Tom Mattson

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Rizzamabob said:Thanks, atm im working this way

OK, now we're cooking.

I think that it would behoove you to either work out or look up the derivative of the hyperbolic cosine function. You'll need it later.

y = (x^2 (e^x + e^-x))/2 + ([itex]\log_e(x)[/itex](e^x + e^-x))/2

No, that's not it.

The way you've written the problem, it should be the hyperbolic cosine

It should be as follows:

[tex]\cosh\left(x^2+log_e(x)\right)=\frac{e^{\left(x^2+log_e(x)\right)}+e^{-\left(x^2+log_e(x)\right)}}{2}[/tex]

Now, that does look nasty but fortunately you don't have to deal with it as it is written on the right hand side. Once you know the derivative of the hyperbolic cosine, you need only to recognize that you have the following composite function:

[tex]f(u)=\cosh(u)[/tex]

[tex]u(x)=x^2+\log_e(x)[/tex]

Do you know what rule you need to use to differentiate a composite function?

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