Question

If $y=(x+\sqrt{x^2-a^2}{)}^n$ then $(x^2-a^2)(\frac{dy}{dx}{)}^2$=

• A. $n^{2}y$
• B. $-n^{2}y$
• C. $n{y}^2$
• D. $n^2{y}^2$

Answer 1

The correct option is D $n^2{y}^2$

$y=(x+\sqrt{x^2-a^2}{)}^n$

$y^{\prime }=n(x+\sqrt{x^2-a^2}{)}^{n-1}(1+\frac{1\left(2x\right)}{2\sqrt{x^2-a^2}})$

$=n\frac{y}{x+\sqrt{x^2-a^2}}(1+\frac{x}{\sqrt{x^2-a^2}})$

$=n\frac{y}{(x+\sqrt{x^2-a^2})}+\frac{\left(\sqrt{x^2+a^2+x}\right)}{\sqrt{x^2-a^2}}$

$(y^{\prime }{)}^2=(\frac{ny}{x^2-a^2}{)}^2$

$(y^{\prime }{)}^2=\frac{n^2{y}^2}{x^2-a^2}$

$\therefore (x^2-a^2)\left(y^{\prime }\right)=(x^2-a^2)\frac{x^2{y}^2}{(x^2-a^2)}$

$=n^2{y}^2$ option is D is correct.