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The derivatives of the hyperbolic functions are quite straightforward and somewhat analogous to the derivatives of their trigonometric counterparts. Here are the derivatives for the six primary hyperbolic functions:
(1) Hyperbolic Sine $sinh$:
$$
\frac{d}{dx} \sinh(x) = \cosh(x)
$$
The derivative of hyperbolic sine is hyperbolic cosine.
Proof
The hyperbolic sine and cosine functions are defined as follows:
$$
\sinh(x) = \frac{e^x – e^{-x}}{2}
$$
$$
\cosh(x) = \frac{e^x + e^{-x}}{2}
$$
Now,
$$
\frac{d}{dx} \sinh(x) = \frac{d}{dx} \left( \frac{e^x – e^{-x}}{2} \right)
$$
$$
= \frac{1}{2}\left( \frac{d}{dx}(e^x) – \frac{d}{dx}(e^{-x})\right)
$$
$$
= \frac{1}{2}(e^x + e^{-x})
$$
This is the definition of $\cosh(x)$:
$$
= \cosh(x)
$$
(2) Hyperbolic Cosine $cosh$:
$$
\frac{d}{dx} \cosh(x) = \sinh(x)
$$
The derivative of hyperbolic cosine is hyperbolic sine.
Proof
The hyperbolic sine and cosine functions are defined as follows:
$$
\sinh(x) = \frac{e^x – e^{-x}}{2}
$$
$$
\cosh(x) = \frac{e^x + e^{-x}}{2}
$$
Now,
$$
\frac{d}{dx}\cosh(x) = \frac{d}{dx}\left( \frac{e^x + e^{-x}}{2}\right)
$$
$$
= \frac{1}{2}\left( \frac{d}{dx}(e^x) + \frac{d}{dx}(e^{-x})\right)
$$
$$
= \frac{1}{2}(e^x – e^{-x})
$$
This is the definition of $\sinh(x)$:
$$
= \sinh(x)
$$
(3) Hyperbolic Tangent $tanh$:
$$
\frac{d}{dx} \tanh(x))= sech^2(x)
$$
The derivative of hyperbolic tangent is the square of hyperbolic secant.
Proof
$$
\frac{d}{dx}\tanh(x) = \frac{d}{dx}\left( \frac{\sinh(x)}{\cosh(x)}\right) = \frac{\cosh(x) \frac{d}{dx} \sinh(x)) – \sinh(x)\frac{d}{dx} \cosh(x))}{\cosh^2(x)}
$$
We know that $\frac{d}{dx} \sinh(x) = \cosh(x)$ and $\frac{d}{dx} \cosh(x) = \sinh(x)$, so substituting these in gives:
$$
= \frac{\cosh(x)\cosh(x) – \sinh(x)\sinh(x)}{\cosh^2(x)}
$$
Simplifying the numerator:
$$
= \frac{\cosh^2(x) – \sinh^2(x)}{\cosh^2(x)}
$$
Using the identity \$cosh^2(x) – \sinh^2(x) = 1$, we get:
$$
= \frac{1}{\cosh^2(x)}
$$
$$
\frac{d}{dx} \tanh(x)) = sech^2(x)
$$
(4) Hyperbolic Cotangent $coth$:
$$
\frac{d}{dx} \coth(x) = -csch^2(x)
$$
The derivative of hyperbolic cotangent is the negative square of hyperbolic cosecant.
Proof
$$
\frac{d}{dx} \coth(x)= \frac{d}{dx}\left( \frac{\cosh(x)}{\sinh(x)}\right) = \frac{\sinh(x)\frac{d}{dx} \cosh(x)) – \cosh(x)\frac{d}{dx} \sinh(x))}{\sinh^2(x)}
$$
We know that $\frac{d}{dx} \cosh(x)) = \sinh(x)$ and $\frac{d}{dx}\sinh(x)) = \cosh(x)$, so substituting these in gives:
$$
= \frac{\sinh(x)\sinh(x) – \cosh(x)\cosh(x)}{\sinh^2(x)}
$$
$$
= \frac{\sinh^2(x) – \cosh^2(x)}{\sinh^2(x)}
$$
Using the identity $\cosh^2(x) – \sinh^2(x) = 1$, we can rewrite the numerator as -1
$$
= \frac{-1}{\sinh^2(x)}
$$
$$
\frac{d}{dx}$coth(x)) = -csch^2(x)
$$
This completes the proof.
(5) Hyperbolic Secant $sech$:
$$
\frac{d}{dx}(sech(x) = -sech(x)\tanh(x)
$$
The derivative of hyperbolic secant involves both hyperbolic secant and hyperbolic tangent.
Proof
$$
\frac{d}{dx} sech(x) = \frac{d}{dx}\left( \frac{1}{\cosh(x)}\right) = \frac{\cosh(x)(0) – 1 \sinh(x))}{\cosh^2(x)}
$$
$$
= \frac{-\sinh(x)}{\cosh^2(x)}
$$
Now, we can express $\sinh(x)$ in terms of $\tanh(x)$ and $\cosh(x)$.
Recall that $\tanh(x) = \frac{\sinh(x)}{\cosh(x)}$, so $\sinh(x) = \tanh(x) \cdot \cosh(x)$.
Substituting this into our equation:
$$
= \frac{-\tanh(x) \cdot \cosh(x)}{\cosh^2(x)}
$$
$$
= -\tanh(x) \cdot \frac{1}{\cosh(x)}
$$
Since $\frac{1}{\cosh(x)}$ is $sech(x)$, we finally have:
$$
\frac{d}{dx}(sech(x)) = -sech(x)\tanh(x)
$$
(6) Hyperbolic Cosecant $csch$:
$$
\frac{d}{dx}(csch(x) = -csch(x)\coth(x)
$$
The derivative of hyperbolic cosecant involves both hyperbolic cosecant and hyperbolic cotangent.
Proof
$$
\frac{d}{dx} csch(x) = \frac{d}{dx}\left( \frac{1}{\sinh(x)}\right) = \frac{\sinh(x)(0) – 1 \cosh(x))}{\sinh^2(x)}
$$
$$
= \frac{-\cosh(x)}{\sinh^2(x)}
$$
Now, we can express $\cosh(x)$ in terms of $\coth(x)$ and $\sinh(x)$.
Recall that $\coth(x) = \frac{\cosh(x)}{\sinh(x)}$, so $\cosh(x) = \coth(x) \cdot \sinh(x)$.
Substituting this into our equation:
$$
= \frac{-\coth(x) \cdot \sinh(x)}{\sinh^2(x)}
$$
$$
= -\coth(x) \cdot \frac{1}{\sinh(x)}
$$
Since $\frac{1}{\sinh(x)}$ is $csch(x)$, we finally have:
$$
\frac{d}{dx}(csch(x)) = -csch(x)\coth(x)
$$
