W wondering New member Joined May 19, 2013 Messages 7 May 20, 2013 #1 \(\displaystyle y=\frac{e^x-e^{-x}}{2}\) \(\displaystyle =\frac{1}{2}e^x-\frac{1}{2}e^{-x}\) \(\displaystyle \frac{dy}{dx}=\frac{1}{2}e^x+\frac{1}{2}e^x\) \(\displaystyle =e^x\) Is this correct.
\(\displaystyle y=\frac{e^x-e^{-x}}{2}\) \(\displaystyle =\frac{1}{2}e^x-\frac{1}{2}e^{-x}\) \(\displaystyle \frac{dy}{dx}=\frac{1}{2}e^x+\frac{1}{2}e^x\) \(\displaystyle =e^x\) Is this correct.
D DrPhil Senior Member Joined Nov 29, 2012 Messages 1,383 May 20, 2013 #2 wondering said: \(\displaystyle y=\frac{e^x-e^{-x}}{2}\) \(\displaystyle =\frac{1}{2}e^x-\frac{1}{2}e^{-x}\) \(\displaystyle \frac{dy}{dx}=\frac{1}{2}e^x+\frac{1}{2}e^x\) \(\displaystyle =e^x\) Is this correct. Click to expand... No. When you take the derivative of e^{-x}, the sign in front changes but the sign in the exponent does not change. \(\displaystyle \displaystyle y' = \frac{e^x\ +\ e^{-x}}{2}\) BTW, the function y also has the name sinh(x), and its derivative is cosh(x).
wondering said: \(\displaystyle y=\frac{e^x-e^{-x}}{2}\) \(\displaystyle =\frac{1}{2}e^x-\frac{1}{2}e^{-x}\) \(\displaystyle \frac{dy}{dx}=\frac{1}{2}e^x+\frac{1}{2}e^x\) \(\displaystyle =e^x\) Is this correct. Click to expand... No. When you take the derivative of e^{-x}, the sign in front changes but the sign in the exponent does not change. \(\displaystyle \displaystyle y' = \frac{e^x\ +\ e^{-x}}{2}\) BTW, the function y also has the name sinh(x), and its derivative is cosh(x).
