Question:
If $sec\left(\frac{x^{2}-2x}{x^{2}+1}\right)=y$, then $\frac{dy}{dx}$ is equal to
If $sec\left(\frac{x^{2}-2x}{x^{2}+1}\right)=y$, then $\frac{dy}{dx}$ is equal to
Updated On: Oct 15, 2024
- $\frac{y^{2}}{x^{2}}$
- $\frac{2y\sqrt{y^{2}-1}\left(x^{2}+x-1\right)}{\left(x^{2}+1\right)^{2}}$
- $\frac{\left(x^{2}+x-1\right)}{y\sqrt{y^{2}-1}}$
- $\frac{x^{2}-y^{2}}{x^{2}+y^{2}}$
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The Correct Option is B
Solution and Explanation
$y=sec\left(\frac{x^{2}-2x}{x^{2}+1}\right)$
$\Rightarrow \frac{dy}{dx}=sec\left(\frac{x^{2}-2x}{x^{2}+1}\right)tan\left(\frac{x^{2}-2x}{x^{2}+1}\right)$.
$\left\{\frac{\left(x^{2}+1\right)\left(2x-2\right)-\left(x^{2}-2x\right)\left(2x\right)}{\left(x^{2}+1\right)^{2}}\right\}$
$=sec\left(\frac{x^{2}-2x}{x^{2}+1}\right)tan\left(\frac{x^{2}-2x}{x^{2}+1}\right)\cdot\frac{2x^{2}+2x-2}{\left(x^{2}+1\right)^{2}}$
$\Rightarrow \frac{dy}{dx}=\frac{2y\sqrt{y^{2}-1}\left(x^{2}+x-1\right)}{\left(x^{2}+1\right)^{2}}$
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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.