JEE Mains · Maths · STD 12 - 5. continuity and differentiation
If \(x^2 + y^2 + sin\, y = 4\), then the value of \(\frac{{{d^2}y}}{{d{x^2}}}\) at the point \((- 2, 0)\) is
- A \(-34\)
- B \(-32\)
- C \(-2\)
- D \(4\)
Answer & Solution
Correct Answer
(A) \(-34\)
Step-by-step Solution
Detailed explanation
Given, \({x^2} + {y^2} + \sin y = 4\) After differentiating the above equation w.r.t.\(x\) we get \(2x + 2y + \frac{{dy}}{{dx}} + \cos y\frac{{dy}}{{dx}} = 0\,......\left( 1 \right)\) \( \Rightarrow 2x + \left( {2y + \cos y} \right)\frac{{dy}}{{dx}} = 0\)…
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