Question

If $y = x\, tan \,x$, then $dy/dx$ is equal to

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

Answer 1

Answer: (D)

We have, $y = xtanx$
$\therefore \frac{dy}{dx}=x\left(tan\,x\right)'+tan\,x\cdot\left(x\right)'$
$=x\,sec^{2}\,x+tan\,x\cdot1$
$=\frac{x}{cos^{2}\,x}+\frac{sin\,x}{cos\,x}=\frac{x+sin\,x\,cos\,x}{cos^{2}\,x}$