Question:
If $y = \tan \: x $ then $ \frac{d^2 y}{dx^2} $ =
If $y = \tan \: x $ then $ \frac{d^2 y}{dx^2} $ =
Updated On: Jun 21, 2022
- $ 1 + y^2$
- $2y ( 1 + y^2)$
- $y( 1 + y^2)$
- $2y(1 -y^2)$
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The Correct Option is B
Solution and Explanation
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)$
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)$
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Concepts Used:
Continuity & Differentiability
Definition of Differentiability
f(x) is said to be differentiable at the point x = a, if the derivative f ‘(a) be at every point in its domain. It is given by
Definition of Continuity
Mathematically, a function is said to be continuous at a point x = a, if
It is implicit that if the left-hand limit (L.H.L), right-hand limit (R.H.L), and the value of the function at x=a exist and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is unspecified or does not exist, then we say that the function is discontinuous.