If for $x \epsilon \left(0, \frac{1}{4}\right) ,$ the derivative of $ \tan^{-1} \left(\frac{6x \sqrt{x}}{1-9x^{3}}\right) $ is $\sqrt{x} . g(x)$, then $g(x)$ equals :
- $\frac{3x \sqrt{x}}{1 - 9x^3}$
- $\frac{3x }{1 - 9x^3}$
- $\frac{3 }{1 + 9x^3}$
- $\frac{9 }{1 + 9x^3}$
The Correct Option is D
Solution and Explanation
For x $\in\left(0, \frac{1}{4}\right)$
$f'\left(x\right) = \frac{9\sqrt{x}}{1+9x^{3}}$
$g\left(x\right) = \frac{9}{1+9x^{3}}$
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Concepts Used:
Continuity
A function is said to be continuous at a point x = a, if
limx→a
f(x) Exists, and
limx→a
f(x) = f(a)
It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is undefined or does not exist, then we say that the function is discontinuous.
Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:
- The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
- The limit of the function as x approaches a, exists.
- The limit of the function as x approaches a, must be equal to the function value at x = a.