Question

If $y=e^{\sqrt{x}}+e^{-\sqrt{x}}$ then $\frac{dy}{dx}$ equals

• A. $\frac{e^{\sqrt{x}}-e^{-\sqrt{x}}}{2\sqrt{x}}$
• B. $\frac{e^{\sqrt{x}}-e^{-\sqrt{x}}}{2x}$
• C. $\frac{1}{2\sqrt{x}}\sqrt{y^2-4}$
• D. $\frac{1}{2\sqrt{x}}\sqrt{y^2+4}$

Answer 1

Answer: (A), (C)

$\frac{e^{\sqrt{x}}-e^{-\sqrt{x}}}{2\sqrt{x}}$
$\frac{1}{2\sqrt{x}}\sqrt{y^2-4}$

$y=e^{\sqrt{x}}+e^{-\sqrt{x}}\to \left(1\right)$

$\frac{dy}{dx}=e^{\sqrt{x}}\frac{1}{2\sqrt{x}}-e^{-\sqrt{x}}\frac{1}{2\sqrt{x}}=\frac{1}{2\sqrt{x}}(e^{\sqrt{x}}-e^{-\sqrt{x}})$

(1) can also be written as, $y^2=(e^{\sqrt{x}}+e^{-\sqrt{x}}{)}^2=(e^{\sqrt{x}}-e^{-\sqrt{x}}{)}^2+4\left(e^{\sqrt{x}}{e}^{-\sqrt{x}}\right)$

$\Rightarrow y^2=(e^{\sqrt{x}}+e^{-\sqrt{x}}{)}^2+4$

$\therefore \frac{dy}{dx}=\frac{1}{2\sqrt{x}}\sqrt{y^2-4}$