## Calculus: Early Transcendentals 8th Edition

a) $$\frac{e^8-1}{2e^4}$$ b) $$\frac{15}{8}$$
Recall that: $$\sinh x =\frac{e^x-e^{-x}}{2}$$ a) $\sinh 4$ $=\frac{e^4-e^{-4}}{2} =\frac{e^4-\frac{1}{e^4}}{2} =\frac{\frac{e^8-1}{e^4}}{2} =\frac{e^8-1}{2e^4}$ b) $\sinh ({\ln4})$ $=\frac{e^{\ln 4}-e^{-({\ln 4})}}{2} =\frac{4-\frac{1}{4}}{2} =\frac{\frac{15}{4}}{2} =\frac{15}{8}$