Question

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

• A. $\frac{e^{\sqrt{x}}-e^{-\sqrt{x}}}{2 \sqrt{x}}$
• B. $\frac{e^{\sqrt{x}}-e^{-\sqrt{x}}}{2 x}$
• 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)

Correct answer is (a) $\frac{e^{\sqrt{x}}-e^{-\sqrt{x}}}{2 \sqrt{x}}$Correct answer is (c) $\frac{1}{2 \sqrt{ x }} \sqrt{ y ^2-4}$