If y = {left[ {x + sqrt {{x^2} - 1} } right]^ | vedclass

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(A) Given $y = {\left( {x + \sqrt {{x^2} - 1} } \right)^{15}} + {\left( {x - \sqrt {{x^2} - 1} } \right)^{15}}$.Differentiating with respect to $x$:$\

Vedclass · Accepted Answer

$225y$

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(A) Given $y = {\left( {x + \sqrt {{x^2} - 1} } \right)^{15}} + {\left( {x - \sqrt {{x^2} - 1} } \right)^{15}}$.Differentiating with respect to $x$:$\frac{{dy}}{{dx}} = 15{\left( {x + \sqrt {{x^2} - 1} } \right)^{14}}\left( {1 + \frac{x}{{\sqrt {{x^2} - 1} }}} \right) + 15{\left( {x - \sqrt {{x^2} - 1} } \right)^{14}}\left( {1 - \frac{x}{{\sqrt {{x^2} - 1} }}} \right)$$\frac{{dy}}{{dx}} = 15{\left( {x + \sqrt {{x^2} - 1} } \right)^{14}}\left( {\frac{{\sqrt {{x^2} - 1} + x}}{{\sqrt {{x^2} - 1} }}} \right) - 15{\left( {x - \sqrt {{x^2} - 1} } \right)^{14}}\left( {\frac{{x - \sqrt {{x^2} - 1} }}{{\sqrt {{x^2} - 1} }}} \right)$$\frac{{dy}}{{dx}} = \frac{{15}}{{\sqrt {{x^2} - 1} }}\left[ {{{\left( {x + \sqrt {{x^2} - 1} } \right)}^{15}} - {{\left( {x - \sqrt {{x^2} - 1} } \right)}^{15}}} \right]$.Note that this approach is slightly complex,so let $u = x + \sqrt{x^2-1}$ and $v = x - \sqrt{x^2-1}$. Then $uv = 1$ and $y = u^{15} + v^{15}$.$\frac{dy}{dx} = 15u^{14} \frac{du}{dx} + 15v^{14} \frac{dv}{dx}$. Since $\frac{du}{dx} = \frac{u}{\sqrt{x^2-1}}$ and $\frac{dv}{dx} = \frac{-v}{\sqrt{x^2-1}}$,we get $\frac{dy}{dx} = \frac{15}{\sqrt{x^2-1}}(u^{15} - v^{15})$.Let $y_1 = u^{15} + v^{15}$ and $y_2 = u^{15} - v^{15}$. Then $\sqrt{x^2-1} \frac{dy}{dx} = 15 y_2$.Differentiating again: $\frac{x}{\sqrt{x^2-1}} \frac{dy}{dx} + \sqrt{x^2-1} \frac{d^2y}{dx^2} = 15 \frac{dy_2}{dx}$.Since $\frac{dy_2}{dx} = 15(u^{14} \frac{du}{dx} - v^{14} \frac{dv}{dx}) = \frac{15}{\sqrt{x^2-1}}(u^{15} + v^{15}) = \frac{15y}{\sqrt{x^2-1}}$.Multiplying by $\sqrt{x^2-1}$: $x \frac{dy}{dx} + (x^2-1) \frac{d^2y}{dx^2} = 15(15y) = 225y$.

If $y = {\left[ {x + \sqrt {{x^2} - 1} } \right]^{15}} + {\left[ {x - \sqrt {{x^2} - 1} } \right]^{15}}$,then $\left( {{x^2} - 1} \right)\frac{{{d^2}y}}{{d{x^2}}} + x\frac{{dy}}{{dx}}$ is equal to

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