exponential derivative

[wondering]

\(\displaystyle y=\frac{e^x-e^{-x}}{2}\)

\(\displaystyle =\frac{1}{2}e^x-\frac{1}{2}e^{-x}\)

\(\displaystyle \frac{dy}{dx}=\frac{1}{2}e^x+\frac{1}{2}e^x\)

\(\displaystyle =e^x\)

Is this correct.

 

[DrPhil]

\(\displaystyle y=\frac{e^x-e^{-x}}{2}\)

\(\displaystyle =\frac{1}{2}e^x-\frac{1}{2}e^{-x}\)

\(\displaystyle \frac{dy}{dx}=\frac{1}{2}e^x+\frac{1}{2}e^x\)

\(\displaystyle =e^x\)

Is this correct.

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]

Thank you.