Question

If $y = \tan \: x$ then $\frac{d^2 y}{dx^2}$ =

• A. $1 + y^2$
• B. $2y ( 1 + y^2)$
• C. $y( 1 + y^2)$
• D. $2y(1 -y^2)$

Answer 1

Answer: (B)

$2y ( 1 + y^2)$

Answer 2

Given, $y = \tan \: x$
Differentiating w.r.t. $'x'$ on both sides $\frac{dy}{dx} = sec^2 \, x$
Taking again derivative w.r.t. $'x' \:\:\:\frac{d^2 y}{dx^2} =2 sec \, x . \sec \, x \, \tan \, x$
$= 2sec^2 x \, \tan\, x = 2\, \tan \, x (1 + tan^2 \, x)$
$= 2y (1 + y^2)$