y + sin-1 (1 - x2) = ex, అయితే, అపుడు  

  1. \(\rm e^x - {2\over\sqrt{2-x^2}}\)
  2. \(\rm e^x + {2\over\sqrt{2-x^2}}\)
  3. \(\rm e^x - {1\over\sqrt{2+x^2}}\)
  4. \(\rm e^x + {1\over\sqrt{2-x^2}}\)

Answer (Detailed Solution Below)

Option 2 :
\(\rm e^x + {2\over\sqrt{2-x^2}}\)
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Detailed Solution

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లెక్కింపు: 

y + sin-1 (1 - x2) = ex 

y = ex - sin-1 (1 - x2

w.r.t x, బేధం ద్వారా మనం పొందేదీ 

\(\rm {dy\over dx} = e^x - {1\over\sqrt{1-(1-x^2)^2}}(-2x)\)

\(\rm {dy\over dx} = e^x + {2x\over\sqrt{1-(1-2x^2+x^4)}}\)

\(\rm {dy\over dx} = e^x + {2x\over\sqrt{2x^2-x^4}}\)

\(\boldsymbol{\rm {dy\over dx} = e^x + {2\over\sqrt{2-x^2}}}\)

 

 

 

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