Chapter 4.1, Problem 14E

To determine

To calculate: The derivative of y with respect to x .

Answer to Problem 14E

  $\frac{dy}{dx}=\frac{\csc x}{\csc x+\cot x}$

Explanation of Solution

Given information: The equation of y in terms of x is given as,

  $y={\left(\csc x+\cot x\right)}^{-1}$

Calculation: Using the derivation,

  $\begin{array}{l}\frac{dy}{dx}=\frac{d{\left(\csc x+\cot x\right)}^{-1}}{dx}\\ \frac{dy}{dx}=-{\left(\csc x+\cot x\right)}^{-2}\times \frac{d\left(\csc x+\cot x\right)}{dx}\\ \frac{dy}{dx}=-{\left(\csc x+\cot x\right)}^{-2}\times \left(\frac{d\left(\csc x\right)}{dx}+\frac{d\left(\cot x\right)}{dx}\right)\\ \frac{dy}{dx}=-{\left(\csc x+\cot x\right)}^{-2}\times \left(-\left(\csc x\right)\left(\cot x\right)-{\csc }^{2}x\right)\\ \frac{dy}{dx}=\frac{-\csc x\left(\csc x+\cot x\right)}{-{\left(\csc x+\cot x\right)}^2}\\ \frac{dy}{dx}=\frac{\csc x}{\csc x+\cot x}\end{array}$

So the derivative of y with respect to x is $\frac{dy}{dx}=\frac{\csc x}{\csc x+\cot x}$