Question:
If $y^{2}=100 \tan^{-1}x+45 sec^{-1}x ,$ then $\frac{dy}{dx}=$
If $y^{2}=100 \tan^{-1}x+45 sec^{-1}x ,$ then $\frac{dy}{dx}=$
Updated On: Jun 7, 2024
- $\frac{x^{2}-1}{x^{2}+1}$
- $\frac{x^{2}+1}{x^{2}-1}$
- 1
- $0$
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The Correct Option is D
Solution and Explanation
$ y^{2}= 100 \tan ^{-1} x+45 \sec ^{-1} x $
$+100 \cot ^{-1} x+45 \operatorname{cosec}^{-1} x $
$= 100 \tan ^{-1} x+100 \cot ^{-1} x $
$ +45 \sec ^{-1} x+45 \operatorname{cosec}^{-1} x $
$= 100\left(\tan ^{-1} x+\cot ^{-1} x\right) $
$+45\left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right) $
$= 100 \times \frac{\pi}{2}+45 \times \frac{\pi}{2} $
On differentiating both sides w.r.t. $X$, we get
$2 y y^{\prime}=0$
$\Rightarrow y^{\prime}=0 [\because 2 \neq 0, y \neq 0]$
$+100 \cot ^{-1} x+45 \operatorname{cosec}^{-1} x $
$= 100 \tan ^{-1} x+100 \cot ^{-1} x $
$ +45 \sec ^{-1} x+45 \operatorname{cosec}^{-1} x $
$= 100\left(\tan ^{-1} x+\cot ^{-1} x\right) $
$+45\left(\sec ^{-1} x+\operatorname{cosec}^{-1} x\right) $
$= 100 \times \frac{\pi}{2}+45 \times \frac{\pi}{2} $
On differentiating both sides w.r.t. $X$, we get
$2 y y^{\prime}=0$
$\Rightarrow y^{\prime}=0 [\because 2 \neq 0, y \neq 0]$
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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.