Question

If $y=e^{\sqrt{x}}+e^{-\sqrt{x}}$, then $\frac{dy}{dx}$ is equal to

• 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)

The correct options are
A $\frac{e^{\sqrt{x}}-e^{-\sqrt{x}}}{2\sqrt{x}}$
C $\frac{1}{2\sqrt{x}}\sqrt{y^2-4}$

We have,
$y=e^{\sqrt{x}}+e^{-\sqrt{x}}$
Differentiate w.r. to $x$
$\frac{dy}{dx}=\frac{e^{\sqrt{x}}}{2\sqrt{x}}-\frac{e^{-\sqrt{x}}}{2\sqrt{x}}=\frac{e^{\sqrt{x}}-e^{-\sqrt{x}}}{2\sqrt{x}}$ ....................option (A)
We can write $e^{\sqrt{x}}-e^{-\sqrt{x}}=\sqrt{{(e^{\sqrt{x}}+e^{-\sqrt{x}})}^2-4}$
$\therefore \frac{dy}{dx}=\frac{\sqrt{{(e^{\sqrt{x}}+e^{-\sqrt{x}})}^2-4}}{2\sqrt{x}}=\frac{\sqrt{y^2-4}}{2\sqrt{x}}$ .................option (C)