Chapter A6, Problem 23E

To determine

To find: the derivative of the given equation of y with respect to appropriate variable.

Answer to Problem 23E

The derivative is $\frac{dy}{dx}=2$ .

Explanation of Solution

Given:

  $y=(x^2+1)sech(lnx)$

Calculation:

The derivative of y with respect to $x$ is calculated below

  $y=(x^2+1)sech(lnx)$

Differentiating y with respect to $x$

  $\begin{array}{l}y=(x^2+1)sech(lnx)\\ y=(x^2+1)(\frac{2}{e^{\ln x}+e^{-\ln x}})\\ y=(x^2+1)(\frac{2}{x+x^{-1}})\\ y=(x^2+1)(\frac{2x}{x^2+1})\\ y=2x\\ \frac{dy}{dx}=2\end{array}$

Conclusion:

Therefore, the derivative is $\frac{dy}{dx}=2$ .