It is given that , [tex]5 x^{2} + y ^{4}= 21\\\\y^{4}= 21 - 5 x^{2}[/tex]
Differentiating both sides with respect to x, we get
→4 y³ y' = 0 - 10 x , where y' is first derivative of y w.r.t x.
→4 y³ y' = - 10 x
When x=2, y=1, substituting these values in above equation
→4× 1³×y'= -10×2
→ y'= -20/4= -5
Differentiating again w.r.t x,the above equation we get
→4 [ 3 y²y'² + y³ y''] = -10 where y'' is double derivative of y w.r.t x.
→3 y²y'² + y³ y'' = -10/4= -5/2
Substituting x=2, y=1, y'= -5,in above equation we get
→ 3 ×1²×(-5)² + 1³× y'' = -5/2
→ 75 + y'' = -5/2
→ y''= -5/2 - 75
→ y'' = [tex]\frac{-150-5}{2}=\frac{-155}{2}= -77.50[/tex]
→ y''= -77.50(Answer)
