ERTW

Initially, I had the identity for sinh at the top (e^x - e^-x)/2 This is my first time with a forum, but i've seen people log their edits, so I figured I should too. In hindsight, writing that didn't really clarify anything and just led to more confusion. my bad. Thanks for confirming my solution by the way, that was one question I felt a little shaky on.

nothing!

Hi all, first post On my calculus exam we had the following question: find the arc length of [latex] \frac{e^x+e^{-x}}{2}[/latex] from zero to a I subbed in hyperbolic cosine and found the arc length to be sinh(a) Any thoughts? Thanks. I only ask because the arc length was crucial to later questions involving area of rotation and none of my classmates used the hyperbolic cosine substitution My solution (don't laugh): [math](\frac{dy}{dx})^2=sinh^2(x)[/math] [math]L = \int_{0}^{a} \sqrt{1+sinh^2(x)}dx[/math] [math]L = \int_{0}^{a} \sqrt{cosh^2(x)}dx[/math] [math]L = \int_{0}^{a} cosh(x)dx[/math] [math]L = sinh(a) - sinh(0) = sinh(a) [/math] edit: e^x + e^-x

you want to take the derivative of that function? if that is what you want, its simple power rule: [math] \frac{df(x)}{dx} = (-3)(-8)x^{-8-1} + 2\frac{1}{2}x^{1/2-1} [/math] you get: [math] \frac{df(x)}{dx} = \frac{24}{x^9} + \frac{1}{\sqrt{x}} [/math]