17.4 Practice Problems
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Solution. Differentiate implicitly, remembering \(\frac {d}{dx}\cosh u=\sinh u\cdot \frac {du}{dx}\) and \(\frac {d}{dx}\sinh x=\cosh x\): \[\sinh y\,\frac {dy}{dx}-12\sinh ^{2}x\cosh x=0 ,\] the second term by the chain rule on \(\sinh ^{3}x\). Hence \[\frac {dy}{dx}=\frac {12\sinh ^{2}x\cosh x}{\sinh y},\quad \sinh y\neq 0 .\]
Note 17.7. The derivatives of \(\sinh \) and \(\cosh \) mirror those of \(\sin \) and \(\cos \) but without the sign change: \[\frac {d}{dx}\sinh x=\cosh x,\qquad \frac {d}{dx}\cosh x=\sinh x .\] Compare \(\frac {d}{dx}\cos x=-\sin x\). The missing minus is the same phenomenon as the sign differences in the identities — Osborn’s rule again, and the reason is the same: the hyperbolic functions are the trigonometric ones with \(x\) replaced by \(ix\).
These follow immediately from the definitions: \(\frac {d}{dx}\cdot \frac {e^{x}-e^{-x}}{2}=\frac {e^{x}+e^{-x}}{2}\), which is \(\cosh x\).
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