Question

If $y = x\, tan\, y$, then $\frac{dy}{dx}=$

• A. $\frac{tan\,x}{x-x^{2}-y^{2}}$
• B. $\frac{y}{x-x^{2}-y^{2}}$
• C. $\frac{tan\,y}{y-x}$
• D. $\frac{tan\,x}{x-y^{2}}$

Answer 1

Answer: (B)

Given $y = x \,tan \,y \,\ldots (i)$
$\Rightarrow \frac{dy}{dx}=x\,sec^{2}\,y \frac{dy}{dx}+tan\,y$
$\Rightarrow \frac{dy}{dx}=\frac{-tan\,y}{x\,sec^{2}\,y-1}$
$=\frac{tan\,y}{1-x\,sec^{2}\,y}$
$=\frac{tan\,y}{1-x\left(1+tan^{2}\,y\right)}$
$=\frac{\frac{y}{x}}{1-x\left(1+\left(\frac{y}{x}\right)^{2}\right)}$ (Using $(i)$
$=\frac{y}{x-x^{2}-y^{2}}$