Question

If $y=sec{x}^0$, then $\frac{dy}{dx}=$

• A. $sec\left(x\right)\tan \left(x\right)$
• B. $sec\left(x^0\right)\tan \left(x^0\right)$
• C. $sec\left(x^0\right)\tan \left(x^0\right)\frac{\mathrm{\pi }}{180}$
• D. $sec\left(x^0\right)\tan \left(x^0\right)\frac{180}{\mathrm{\pi }}$

Answer 1

The correct option is C

$sec\left(x^0\right)\tan \left(x^0\right)\frac{\mathrm{\pi }}{180}$

Explanation for the correct answer:

To find the value of $\frac{dy}{dx}$:

Given,

$y=secx°$

We know,

$\mathrm{\pi }$ radian $=180°$

Then,

$x°=\frac{\mathrm{\pi }}{180}x$

So,

$$\begin{gathered} y=sec\frac{\mathrm{\pi }}{180}x \\ \frac{dy}{dx}=sec\left(\frac{\mathrm{\pi }}{180}x\right)\tan \left(\frac{\mathrm{\pi }}{180}x\right)\frac{\mathrm{\pi }}{180}\mathbf{}\mathbf{[}\mathbf{\because }\mathbf{}\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{}\mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{s}\mathbf{e}\mathbf{c}\mathbf{}\mathbf{x}\mathbf{}\mathbf{t}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{x}\mathbf{]} \\ \frac{dy}{dx}\mathbf{}=sec\left(x°\right)\tan \left(x°\right)\frac{\mathrm{\pi }}{180} \end{gathered}$$

Hence, the correct option is C.