Chapter 4, Problem 30RE

To determine

To Find: The derivative of the given function.

Answer to Problem 30RE

The derivative is $2\left(\frac{1+\sin \theta }{1-\cos \theta }\right)\left(\frac{\cos \theta -\sin \theta -1}{{(1-\cos \theta )}^2}\right)$ .

Explanation of Solution

Given information: Given function $r={\left(\frac{1+\sin \theta }{1-\cos \theta }\right)}^2$

Formula used: $\frac{d(\sin x)}{dx}=\cos x,\frac{d(\cos x)}{dx}=-\sin x\text{and}\frac{d(x^n)}{dx}=n{x}^{n-1}$ with chain, quotient, addition and subtraction rule.

Calculation:

  $\begin{array}{l}r={\left(\frac{1+\sin \theta }{1-\cos \theta }\right)}^2\\ \text{Differentiate}\text{with}\text{respect}\text{to}\theta \\ \frac{dy}{d\theta }=\frac{d{\left(\frac{1+\sin \theta }{1-\cos \theta }\right)}^2}{d\left(\frac{1+\sin \theta }{1-\cos \theta }\right)}\times \frac{d{\left(\frac{1+\sin \theta }{1-\cos \theta }\right)}^{}}{d\theta }\\ \frac{dy}{d\theta }=2\left(\frac{1+\sin \theta }{1-\cos \theta }\right)\times \left[\frac{\frac{d(1+\sin \theta )}{d\theta }\times (1-\cos \theta )-(1+\sin \theta )\times \frac{d(1-\cos \theta )}{d\theta }}{{(1-\cos \theta )}^2}\right]\\ \frac{dy}{d\theta }=2\left(\frac{1+\sin \theta }{1-\cos \theta }\right)\left[\frac{\cos \theta \times (1-\cos \theta )-(1+\sin \theta )\times \sin \theta }{{(1-\cos \theta )}^2}\right]\\ \frac{dy}{d\theta }=2\left(\frac{1+\sin \theta }{1-\cos \theta }\right)\left(\frac{\cos \theta -{\cos }^2\theta -\sin \theta -{\sin }^2\theta }{{(1-\cos \theta )}^2}\right)\\ \frac{dy}{d\theta }=2\left(\frac{1+\sin \theta }{1-\cos \theta }\right)\left(\frac{\cos \theta -\sin \theta -1}{{(1-\cos \theta )}^2}\right)\end{array}$