ok so how exactly would I simplify this function sinh(ln(t))
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Originally Posted by mathlete ok so how exactly would I simplify this function sinh(ln(t)) Use the chain rule: $\displaystyle \frac{d}{dt}sinh(ln(t)) = cosh(ln(t)) \cdot \frac{1}{t}$ -Dan
Before takin' the derivative, observe that $\displaystyle \sinh x = \frac{{e^x - e^{ - x} }} {2}.$ Now contemplate $\displaystyle x=\ln t$ and take its derivative.
yeah i know how to take the derivative, but it asks me to simplify the function first...so to do that would i just plug in ln(t) into the (e^x-e^(-x))/2??
Originally Posted by Krizalid Before takin' the derivative, observe that $\displaystyle \sinh x = \frac{{e^x - e^{ - x} }} {2}.$ Now contemplate $\displaystyle x=\ln t$ and take its derivative. Ha! I never saw that. -Dan
Originally Posted by Krizalid Before takin' the derivative, observe that $\displaystyle \sinh x = \frac{{e^x - e^{ - x} }} {2}.$ Now contemplate $\displaystyle x=\ln t$ and take its derivative. Nice, so the derivative is just 1?
Originally Posted by colby2152 Nice, so the derivative is just 1? No. $\displaystyle sinh(ln(x)) = \frac{e^{ln(x)} - e^{-ln(t)}}{2}$ $\displaystyle = \frac{x - \frac{1}{x}}{2}$ So the derivative will be $\displaystyle \frac{1}{2} + \frac{1}{2x^2}$ -Dan
Originally Posted by topsquark No. $\displaystyle sinh(ln(x)) = \frac{e^{ln(x)} - e^{-ln(t)}}{2}$ $\displaystyle = \frac{x - \frac{1}{x}}{2}$ So the derivative will be $\displaystyle \frac{1}{2} + \frac{1}{2x^2}$ -Dan Whoops, gotta watch where I throw those negative signs!
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