Question

If $x=\cos \theta ,y=\sin 5\theta$, then $(1-x^2)d^{2}y/d{x}^2-xdy/dx$ is equal to:

• A. $-5y$
• B. $5y$
• C. $25y$
• D. $-25y$

Answer 1

The correct option is D

$-25y$

Explanation for the correct option.

Step 1: Differentiate $x\mathrm{and}y$ w.r.t $\theta$

$\begin{array}{rcl}x& =& \cos \theta \\ & \Rightarrow & \frac{dx}{d\theta }=-\sin \theta \end{array}$

$\begin{array}{rcl}y& =& \sin 5\theta \\ & \Rightarrow & \frac{dy}{d\theta }=5\cos 5\theta \end{array}$

Step 2: Find First and second derivative.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{{\displaystyle \frac{dy}{d\theta }}}{{\displaystyle \frac{dx}{d\theta }}}\\ & =& -\frac{5\cos 5\theta }{\sin \theta }\end{array}$

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& \frac{d}{d\theta }\left(\frac{dy}{dx}\right)\left(\frac{d\theta }{dx}\right)\\ & =& -5\frac{d}{d\theta }\left(\frac{\cos 5\theta }{\sin \theta }\right)\times \left(-\frac{1}{\sin \theta }\right)\\ & =& \frac{5}{\sin \theta }\times \frac{-5\sin 5\theta \times \sin \theta -\cos 5\theta \times \cos \theta }{{\sin }^2\theta }\end{array}$

Step 3: Find the value of $(1-x^2)d^{2}y/d{x}^2-xdy/dx$

By substituting the values in the given equation we get,

$\begin{array}{rcl}(1-x^2)d^{2}y/d{x}^2-xdy/dx& =& \left(1-{\cos }^2\theta \right)\left(\frac{5}{\sin \theta }\times \frac{-5\sin 5\theta \times \sin \theta -\cos 5\theta \times \cos \theta }{{\sin }^2\theta }\right)-\cos \theta \left(-\frac{5\cos 5\theta }{\sin \theta }\right)\\ & =& {\sin }^2\theta \times \frac{5\left(-5\sin 5\theta \times \sin \theta -\cos 5\theta \times \cos \theta \right)}{{\sin }^3\theta }-\cos \theta \left(-\frac{5\cos 5\theta }{\sin \theta }\right)\\ & =& \frac{-25\sin 5\theta \times \sin \theta }{\sin \theta }-\frac{5\left(\cos 5\theta \times \cos \theta \right)}{\sin \theta }-\cos \theta \left(-\frac{5\cos 5\theta }{\sin \theta }\right)\\ & =& -25\sin 5\theta -5\cos 5\theta cot\theta +5\cos 5\theta cot\theta \\ & =& -25\sin 5\theta \\ & =& -25y\end{array}$

Hence, option D is correct.