Find y' for y = (e^x - e^-x)/2
Find $y'$ for $y = \dfrac{e^{x} - e^{-x}}{2}$.
Explanation of Method
Differentiate the hyperbolic sine written out in exponentials.
Step 1: Differentiate term by term
Using $\bigl(e^{x}\bigr)' = e^{x}$ (Problem 17) and $\bigl(e^{-x}\bigr)' = -e^{-x}$ (Problem 61), the minus signs combine into a plus: \[y' = \frac{1}{2}\Bigl(\bigl(e^{x}\bigr)' - \bigl(e^{-x}\bigr)'\Bigr) = \frac{1}{2}\Bigl(e^{x} - \bigl(-e^{-x}\bigr)\Bigr) = \frac{e^{x}+e^{-x}}{2}.\]
Step 2: Identify the result
The expression obtained is the hyperbolic cosine: $\cosh x = \frac{e^{x}+e^{-x}}{2}$.
Step 3: Verification
\[y = \frac{e^{x}-e^{-x}}{2} = \sinh x \;\Longrightarrow\; \bigl(\sinh x\bigr)' = \cosh x, \text{ the standard rule} \;\;\checkmark\]
\[x = 0:\quad y'(0) = \frac{1+1}{2} = 1; \qquad \frac{y(10^{-5})-y(-10^{-5})}{2\times 10^{-5}} \approx 1.0 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,y' = \dfrac{e^{x}+e^{-x}}{2} = \cosh x\,}\]
Remark
Remark. Differentiating again gives $\bigl(\cosh x\bigr)' = \sinh x$, so $y'' = y$ — the hyperbolic analogue of $y'' = -y$ for the circular sine. Note that $\cosh x \geq 1 \gt 0$, so $\sinh x$ is strictly increasing on all of $\mathbb{R}$.