Question

If $x=sec\theta -cos\theta ,y=se{c}^n\theta -co{s}^n\theta$, then $\frac{dy}{dx}=$

Answer 1

$x=\sec \theta -\cos \theta$

$x^2={\sec }^2\theta +{\cos }^2\theta -2$

$x^2+4={\sec }^2\theta +\cos \theta +2$

$x^2+a(\sec \theta +\cos \theta {)}^2$

$(\sec \theta +\cos \theta )=\sqrt{x^2+4}$

$y={\sec }^n\theta -{\cos }^n\theta$

$y^2={\sec }^{2n}\theta +{\cos }^{2n}\theta -2$

$y^2+4={\sec }^{2n}\theta +{\cos }^{2n}\theta +2$

$y^2+4=({\sec }^n\theta +{\cos }^n\theta {)}^2$

$\sqrt{y^2+4}=({\sec }^n\theta +{\cos }^n\theta )$

$x=\sec \theta -\cos \theta$

$\frac{dx}{d\theta }=[\sec \theta \tan \theta +\sin \theta ]$

$\frac{dx}{d\theta }\sin \theta [\frac{1}{{\cos }^2\theta }+1]$

$y={\sec }^n\theta -{\cos }^n\theta$

$\frac{dy}{d\theta }=n{\sec }^{n-1}\theta \left(\sec \tan \theta \right)-n{\cos }^{n-1}\theta (-\sin \theta )$

$\frac{dy}{d\theta }=\left(n{\sec }^n\tan \theta \right)+(n{\cos }^n\theta +{\cos }^n\theta )$

$\frac{dy}{d\theta }=n\tan \theta [{\sec }^n\theta +{\cos }^n\theta ]$

$\frac{dy}{dx}=\frac{n\tan \theta [{\sec }^n\theta +{\cos }^n\theta ]}{\left(\sin \theta \right)\left[\frac{1+{\cos }^2\theta }{{\cos }^2\theta }\right]}$

$=\frac{n\left(\frac{\sin \theta }{\cos \theta }\right)({\sec }^n+{\sec }^n\theta \theta ){\cos }^2\theta }{\left(\sin \theta \right)(1+{\cos }^2\theta )}$

$=\frac{n\cos \theta ({\sec }^n\theta +{\cos }^n\theta )}{\cos \theta (\frac{1}{\cos \theta }+\cos \theta )}$

$\frac{dy}{dx}=\frac{({\sec }^n\theta +{\cos }^n\theta )}{(\sec \theta +\cos \theta )}=\frac{n\sqrt{y^2+4}}{\sqrt{x^2+4}}$

$\Rightarrow \frac{dy}{dx}=n\sqrt{\frac{y^2+4}{x^2+4}}$