Rewrite the expression in terms of exponentials and simplify the results.
-$10 \cosh ( \ln x ) + 12 \sinh ( \ln x )$
A) $11 x - \frac { 1 } { x }$
B) $11 \mathrm { x }$
C) 0
D) $11 \left( x + \frac { 1 } { x } \right)$

Question

Quizplus · Accepted Answer

The expression can be simplified using the definitions of hyperbolic sine and cosine in terms of exponentials: $\cosh(y) = \frac{e^y + e^{-y}}{2}$ and $\sinh(y) = \frac{e^y - e^{-y}}{2}$. Substituting $y = \ln(x)$ gives $\cosh(\ln(x)) = \frac{x + \frac{1}{x}}{2}$ and $\sinh(\ln(x)) = \frac{x - \frac{1}{x}}{2}$. Plugging these into the original expression yields $10\left(\frac{x + \frac{1}{x}}{2}\right) + 12\left(\frac{x - \frac{1}{x}}{2}\right) = 5(x + \frac{1}{x}) + 6(x - \frac{1}{x}) = 11x - \frac{1}{x}$.

Rewrite the Expression in Terms of Exponentials and Simplify the Results