# Thread: chain rule issue with derivative involving ln and e

1. ## chain rule issue with derivative involving ln and e

y = ln [(e^x + e^-x) / 2] We need to find the derivative of this.

So I extract the 1/2 so it becomes

y = 1/2 ln (e^x + e^-x)

I'm not sure how the chain rule is applied here. The final answer is; (e^x - e^-x) / (e^x + e^-x)

2. Originally Posted by Archduke01
y = ln [(e^x + e^-x) / 2] We need to find the derivative of this.

So I extract the 1/2 so it becomes

y = 1/2 ln (e^x + e^-x)

I'm not sure how the chain rule is applied here. The final answer is; (e^x - e^-x) / (e^x + e^-x)
"extract" the 1/2?? And exactly what rule says we can do that?!

$\displaystyle y = \ln \frac {e^x + e^{-x}}2$

$\displaystyle = \ln (e^x + e^{-x}) - \ln 2$

Now differentiate using the rule: $\displaystyle \frac d{dx} \ln f(x) = \frac {f'(x)}{f(x)}$

3. Originally Posted by Archduke01
y = ln [(e^x + e^-x) / 2] We need to find the derivative of this.

So I extract the 1/2 so it becomes

y = 1/2 ln (e^x + e^-x)

I'm not sure how the chain rule is applied here. The final answer is; (e^x - e^-x) / (e^x + e^-x)
You can't 'extract' the 2. It is associated with the Logarithm. You must be careful

$\displaystyle y=\ln(e^x-e^{-x})-\ln2$

Now

Let $\displaystyle u=e^x-e^{-x}$

Then

$\displaystyle y'=\frac{u'}{u}=\frac{e^x+e^{-x}}{e^x-e^{-x}}=\coth{x}$

4. Originally Posted by VonNemo19
You can't 'extract' the 2. It is associated with the Logarithm. You must be careful

$\displaystyle y=\ln(e^x-e^{-x})-\ln2$

Now

Let $\displaystyle u=e^x-e^{-x}$

Then

$\displaystyle y'=\frac{u'}{u}=\frac{e^x+e^{-x}}{e^x-e^{-x}}=\coth{x}$
It's $\displaystyle e^x + e^{-x}$

@OP: If we wanted hyperbolic functions for the answer, we could have replaced $\displaystyle \frac {e^x + e^{-x}}2$ with $\displaystyle \cosh x$ from the get go

5. Originally Posted by Jhevon
"extract" the 1/2?? And exactly what rule says we can do that?!

$\displaystyle y = \ln \frac {e^x + e^{-x}}2$

$\displaystyle = \ln (e^x + e^{-x}) - \ln 2$

Now differentiate using the rule: $\displaystyle \frac d{dx} \ln f(x) = \frac {f'(x)}{f(x)}$
if c is a constant then d ca/dx = c * d a/dx

But my badness, seems like I misunderstood the rule.

I don't understand the coth x cosh x stuff; I don't think we've learned that yet.

6. Originally Posted by Jhevon
It's $\displaystyle e^x + e^{-x}$ Oops

@OP: If we wanted hyperbolic functions for the answer, we could have replaced $\displaystyle \frac {e^x + e^{-x}}2$ with $\displaystyle \cosh x$ from the get go I know this. I was trying to get a "Hey, what is coth,?" out of the OP
Thanks for the correction.

7. $\displaystyle = \ln (e^x + e^{-x}) - \ln 2$

how did ln 1/2 become - ln 2?

8. Originally Posted by Archduke01
$\displaystyle = \ln (e^x + e^{-x}) - \ln 2$$\displaystyle how did ln 1/2 become - ln 2? From the basic log rule \displaystyle \ln\frac{a}{b}=\ln{a}-\ln{b}. 9. Originally Posted by VonNemo19 From the basic log rule \displaystyle \ln\frac{a}{b}=\ln{a}-\ln{b}. Thanks bromosapien. \displaystyle y'=\frac{u'}{u}=\frac{e^x+e^{-x}}{e^x-e^{-x}}=\coth{x} Is this from applying the chain rule? How did you guys get rid of the - ln 2? I assume you derived it, getting - 1/2 but at that point I'm stumped. 10. Originally Posted by Archduke01 Thanks bromosapien. \displaystyle y'=\frac{u'}{u}=\frac{e^x+e^{-x}}{e^x-e^{-x}}=\coth{x} Is this from applying the chain rule? How did you guys get rid of the - ln 2? I assume you derived it, getting - 1/2 but at that point I'm stumped. ln(2) is a constant.... 11. Originally Posted by Jhevon ln(2) is a constant.... Ah yes of course, I was just - Thanks for the help, gentlemen. Sorry for the whole back and forth. 12. Originally Posted by Archduke01 Thanks bromosapien. \displaystyle y'=\frac{u'}{u}=\frac{e^x+e^{-x}}{e^x-e^{-x}}=\coth{x}$$\displaystyle$

Is this from applying the chain rule?

How did you guys get rid of the - ln 2? I assume you derived it, getting - 1/2 but at that point I'm stumped.
$\displaystyle \ln2$ is a constant. Constants dissapear when differentiating.

Regarding the chain rule...

You know that $\displaystyle \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$ right?

Well, like I said before (if you forget about that nasty sign mistake that I made), let $\displaystyle u=e^x+e^{-x}$, then by the chain rule we have

$\displaystyle \frac{dy}{dx}=\frac{dy}{du}{\frac{du}{dx}}=\frac{1 }{u}\frac{du}{dx}=\frac{1}{e^x+e^{-x}}\cdot(e^x-e^{-x})$.

See what I'm sayin' brominator?