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)
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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?! Now differentiate using the rule:
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 Now Let Then
Originally Posted by VonNemo19 You can't 'extract' the 2. It is associated with the Logarithm. You must be careful Now Let Then It's @OP: If we wanted hyperbolic functions for the answer, we could have replaced with from the get go
Originally Posted by Jhevon "extract" the 1/2?? And exactly what rule says we can do that?! Now differentiate using the rule: 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.
Originally Posted by Jhevon It's Oops @OP: If we wanted hyperbolic functions for the answer, we could have replaced with 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.
how did ln 1/2 become - ln 2?
Originally Posted by Archduke01 how did ln 1/2 become - ln 2? From the basic log rule
Originally Posted by VonNemo19 From the basic log rule Thanks bromosapien. 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.
Originally Posted by Archduke01 Thanks bromosapien. 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....
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.
Originally Posted by Archduke01 Thanks bromosapien. 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. is a constant. Constants dissapear when differentiating. Regarding the chain rule... You know that right? Well, like I said before (if you forget about that nasty sign mistake that I made), let , then by the chain rule we have . See what I'm sayin' brominator?
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