Question:
If \(y = \sqrt{x+\sqrt{x^2+1}}\), then the value of \(\frac{dy}{dx}\) is
If \(y = \sqrt{x+\sqrt{x^2+1}}\), then the value of \(\frac{dy}{dx}\) is
Show Hint
Square the equation, differentiate, and keep $y$ in the answer.
Updated On: Oct 1, 2026
- \(\frac{\sqrt{x^2+1}+x}{2y\sqrt{x^2+1}}\)
- \(\frac{\sqrt{x^2+1}+x}{y\sqrt{x^2+1}}\)
- \(\frac{\sqrt{x^2+1}-x}{2y\sqrt{x^2+1}}\)
- \(\frac{\sqrt{x^2+1}+x}{2y\sqrt{x^2-1}}\)
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The Correct Option is A
Solution and Explanation
Step 1: Square both sides
\(y^2=x+\sqrt{x^2+1}\).
Step 2: Differentiate
\(2yy'=1+\frac{x}{\sqrt{x^2+1}}=\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}\).
Step 3: Solve for \(y'\)
\[ y'=\frac{\sqrt{x^2+1}+x}{2y\sqrt{x^2+1}} \]
This is option (A).
Final Answer:
\(\frac{dy}{dx}=\frac{\sqrt{x^2+1}+x}{2y\sqrt{x^2+1}}\), option (A). \[ \boxed{\text{(A)}} \]
\(y^2=x+\sqrt{x^2+1}\).
Step 2: Differentiate
\(2yy'=1+\frac{x}{\sqrt{x^2+1}}=\frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}}\).
Step 3: Solve for \(y'\)
\[ y'=\frac{\sqrt{x^2+1}+x}{2y\sqrt{x^2+1}} \]
This is option (A).
Final Answer:
\(\frac{dy}{dx}=\frac{\sqrt{x^2+1}+x}{2y\sqrt{x^2+1}}\), option (A). \[ \boxed{\text{(A)}} \]
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