Question

The derivative of the function $f\left(x\right)=\frac{x^2}{1-{\sin }^{2}x}$is

• A. Even function
• B. Odd function
• C. Not define
• D. Increasing Function

Answer 1

The correct option is B

Odd function

Explanation for the correct option:

Finding if the given function is even or odd:

$$\begin{gathered} \mathrm{Given}\mathrm{function}\mathrm{is}f\left(x\right)=\frac{x^2}{\left(1-{\sin }^{2}x\right)} \\ \mathrm{Therefore}{\sin }^{2}x+{\cos }^{2}x=1 \\ \Rightarrow {\cos }^{2}x=\left(1-{\sin }^{2}x\right) \\ \mathrm{Therefore}f\left(x\right)=\frac{x^2}{{\cos }^{2}x} \\ \frac{d}{dx}\left(f\left(x\right)\right)=\frac{d}{dx}\left(\frac{x^2}{{\cos }^{2}x}\right) \\ =\frac{{\cos }^{2}x\times {\displaystyle \frac{d}{dx}}\left(x^2\right)-x^2\times {\displaystyle \frac{d}{dx}}\left({\cos }^{2}x\right)}{{\left({\cos }^{2}x\right)}^2} \\ =\frac{{\cos }^{2}x\times 2x-x^2\times 2\cos x\left(-\sin x\right)}{{\cos }^{4}x} \\ =\frac{2x{\cos }^{2}x+2x^2\cos x\sin x}{{\cos }^{4}x} \end{gathered}$$

$$\begin{gathered} f^{\text{'}}\left(x\right)=\frac{2x\cos x+2x^2\sin x}{{\cos }^{3}x}\left[\because \mathrm{Cancelling}\mathrm{cosx}\mathrm{from}\mathrm{numerator}\mathrm{and}\mathrm{dinominator}\right] \\ \mathrm{Now},{\mathrm{f}}^{\text{'}}\left(-\mathrm{x}\right)=\frac{2\times \left(-\mathrm{x}\right)\cos \left(-\mathrm{x}\right)+2\times {\left(-\mathrm{x}\right)}^2\sin \left(-\mathrm{x}\right)}{{\left(\cos \left(-\mathrm{x}\right)\right)}^3} \\ =\frac{-2\mathrm{xcosx}+2{\mathrm{x}}^2\times \left(-\mathrm{sinx}\right)}{{\left(\mathrm{cosx}\right)}^3} \\ \mathrm{Therefore}\left[\cos \left(-\mathrm{x}\right)=\mathrm{cosx}\mathrm{and}\sin (-\mathrm{x})=-\mathrm{sinx}\right] \\ =\frac{-2\mathrm{xcosx}-2{\mathrm{x}}^2\mathrm{sinx}}{{\cos }^3\mathrm{x}} \\ =\frac{-\left(2\mathrm{xcosx}+2{\mathrm{x}}^2\mathrm{sinx}\right)}{{\cos }^3\mathrm{x}} \\ \mathrm{Therefore}{\mathrm{f}}^{\text{'}}\left(-\mathrm{x}\right)=-{\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right) \\ \mathrm{Hence}{\mathrm{f}}^{\text{'}}\left(\mathrm{x}\right)\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{function}. \end{gathered}$$

Therefore, the correct answer is option (B).