Question:
If \(y = \sqrt{x^2+8x+3}\), then the value of \(\frac{d^2y}{dx^2}\) at \(x = 3\) is
If \(y = \sqrt{x^2+8x+3}\), then the value of \(\frac{d^2y}{dx^2}\) at \(x = 3\) is
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Use $y^2=x^2+8x+3$ and differentiate twice.
Updated On: Oct 1, 2026
- \(\frac{-13}{216}\)
- \(\frac{13}{216}\)
- \(\frac{15}{216}\)
- \(\frac{-15}{216}\)
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The Correct Option is A
Solution and Explanation
Step 1: First derivative
\(y^2=x^2+8x+3\), so \(2yy'=2x+8\) and \(y'=\frac{x+4}{y}\).
Step 2: Second derivative
Use the quotient rule: \(y''=\frac{y-(x+4)y'}{y^2}=\frac{y^2-(x+4)^2}{y^3}\).
Now \(y^2-(x+4)^2=x^2+8x+3-x^2-8x-16=-13\), so \(y''=\frac{-13}{y^3}\).
Step 3: Evaluate at \(x=3\)
\(y^2=9+24+3=36\), so \(y=6\) and \(y^3=216\).
\[ y''=\frac{-13}{216} \]
Final Answer:
\(\frac{d^2y}{dx^2}=-\frac{13}{216}\), option (A). \[ \boxed{\text{(A) } -\frac{13}{216}} \]
\(y^2=x^2+8x+3\), so \(2yy'=2x+8\) and \(y'=\frac{x+4}{y}\).
Step 2: Second derivative
Use the quotient rule: \(y''=\frac{y-(x+4)y'}{y^2}=\frac{y^2-(x+4)^2}{y^3}\).
Now \(y^2-(x+4)^2=x^2+8x+3-x^2-8x-16=-13\), so \(y''=\frac{-13}{y^3}\).
Step 3: Evaluate at \(x=3\)
\(y^2=9+24+3=36\), so \(y=6\) and \(y^3=216\).
\[ y''=\frac{-13}{216} \]
Final Answer:
\(\frac{d^2y}{dx^2}=-\frac{13}{216}\), option (A). \[ \boxed{\text{(A) } -\frac{13}{216}} \]
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