Question

$\frac{d}{dx}\sqrt{(se{c}^{2}x+\cos e{c}^{2}x)}=$

• A. $4\cos ec2xcot2x$
• B. $-4\cos ec2xcot2x$
• C. $4\cos ecxcot2x$
• D. None of these

Answer 1

The correct option is B

$-4\cos ec2xcot2x$

Finding the value of

Given, $y=\sqrt{(se{c}^{2}x+\cos e{c}^{2}x)}$

$$\begin{gathered} =\sqrt{\left(\frac{1}{{\cos }^{2}x}+\frac{1}{{\sin }^{2}x}\right)} \\ =\sqrt{\frac{{\sin }^{2}x+{\cos }^{2}x}{{\cos }^{2}x{\sin }^{2}x}} \\ =\frac{1}{\cos x\sin x} \\ =\frac{2}{\sin 2x} \\ =2\cos ec2x \end{gathered}$$

Differentiate it with respect to $x$, we get

$\begin{array}{rcl}\frac{dy}{dx}& =& 2\times -\cos ec2xcot2x\times 2\\ & =& \mathbf{}\mathbf{-}\mathbf{4}\mathbf{}\mathit{c}\mathit{o}\mathit{s}\mathit{e}\mathit{c}\mathbf{}\mathbf{2}\mathit{x}\mathbf{}\mathit{c}\mathit{o}\mathit{t}\mathbf{}\mathbf{2}\mathit{x}\end{array}$

Hence, option (B) is correct.