Question

To find the derivative of the function\(f(t) = cscht(1 - lncscht)\).

Short answer

The derivative of the function \(f(t) = {\mathop{\rm csch}\nolimits} t(1 - \ln {\mathop{\rm csch}\nolimits} t)\)is\((\ln {\mathop{\rm csch}\nolimits} t)({\mathop{\rm csch}\nolimits} t\coth t)\).

Step-by-step solution

Given function

The function \(f(t) = {\mathop{\rm csch}\nolimits} t(1 - \ln {\mathop{\rm csch}\nolimits} t)\)

Concept of the Derivative rules

The Derivative rules used:

(1) The chain rule:\(\frac{{dy}}{{dx}} = \frac{{dy}}{{du}}\frac{{du}}{{dx}}\)

(2) The derivative of the hyperbolic cosine function is,\(\frac{d}{{dx}}(coshx) = sinhx\).

Solve the given expression to general complex number

The derivative of the function \(y = {e^{\cosh 3x}}\) is computed as shown below.

\(\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{e^{\cosh 3x}}} \right)\)

Let \(u = \cosh 3x\).

\(\frac{{dy}}{{dx}} = \frac{d}{{dx}}\left( {{e^\mu }} \right)\)

Apply the chain rule and simplify the terms.

\(\begin{array}{c}\frac{{dy}}{{dx}} = \frac{d}{{du}}\left( {{e^u}} \right)\frac{{du}}{{dx}}\\ = {e^u}\frac{{du}}{{dx}}\end{array}\)

Substitute, \(u = \cosh 3x\).

\(\begin{array}{c}\frac{{dy}}{{dx}} = {e^{\cosh 3x}}\frac{d}{{dx}}(\cosh 3x)\\ = {e^{\cosh 3x}}\frac{d}{{d(3x)}}(\cosh 3x) \cdot \frac{d}{{dx}}(3x)\\ = {e^{\cosh 3x}}\sinh 3x(3)\\ = 3{e^{\cosh 3x}}\sinh 3x\end{array}\)

Therefore, the derivative of the function \(y = {e^{\cosh 3x}}\) is, \(\frac{{dy}}{{dx}} = 3{e^{\cosh 3x}}\sinh 3x\).