Question

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

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

Answer 1

Answer: (B)

Given $y=xtany\dots (i)$
$\Rightarrow \frac{dy}{dx}=xse{c}^{2}y\frac{dy}{dx}+tany$
$\Rightarrow \frac{dy}{dx}=\frac{-tany}{xse{c}^{2}y-1}$
$=\frac{tany}{1-xse{c}^{2}y}$
$=\frac{tany}{1-x\left(1+ta{n}^{2}y\right)}$
$=\frac{\frac{y}{x}}{1-x\left(1+{\left(\frac{y}{x}\right)}^2\right)}$ (Using $(i)$
$=\frac{y}{x-x^2-y^2}$