When proving that d/dx (sinhx) = coshx, is this correct: sinhx = (e^x-e^-x) / 2 d/dx sinhx = (e^x+e^-x)/2 Since /2 is just a constant I can leave it like that, correct? Is there any steps that I migth be missing?
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Hello, As 2 is a constant, yes you can. In general, the derivative of ay, where a is a constant will be a*y'
$\displaystyle sinh(x)=\frac{e^{x}-e^{-x}}{2}$...therefore...$\displaystyle \frac{D[sinh(x)]}{dx}=\frac{1}{2}[e^{x}+e^{-x}]=cosh(x)$
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