(A) Given $e^x = y + \sqrt{y^2 - 1}$.Rearranging the terms, we get $e^x - y = \sqrt{y^2 - 1}$.Squaring both sides, we have $(e^x - y)^2 = y^2 - 1$.Exp

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(A) Given $e^x = y + \sqrt{y^2 - 1}$.Rearranging the terms, we get $e^x - y = \sqrt{y^2 - 1}$.Squaring both sides, we have $(e^x - y)^2 = y^2 - 1$.Expanding the square: $e^{2x} + y^2 - 2ye^x = y^2 - 1$.Simplifying, we get $e^{2x} + 1 = 2ye^x$, which implies $y = \frac{e^{2x} + 1}{2e^x} = \frac{e^x + e^{-x}}{2} = \cosh x$.Now, differentiating $y = \cosh x$ with respect to $x$, we get $\frac{dy}{dx} = \sinh x$.

Class 12 Mathematics Continuity and Differentiability Derivative at a point, Standard differentiation

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If $e^{x}=y+\sqrt{y^2-1}$, then $\frac{d y}{d x}=$

TS EAMCET 2023 All TS EAMCET

A
$\sinh x$
B
$\cosh x$
C
$\tanh x$
D
$-\sinh x$

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Derivative at a point, Standard differentiation

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