If `y=sin(m sin^(-1)x)`, what is the value of `d^(2)y//dx^(2)` at x = 0?A. mB. `m^(2)`C. `m^(2)+2`D. None of these

If `y=sin(m sin^(-1)x)`, what is the value of `d^(2)y//dx^(2)` at x =

1.	If `y=sin(m sin^(-1)x)`, what is the value of `d^(2)y//dx^(2)` at x = 0?A. mB. `m^(2)`C. `m^(2)+2`D. None of these
Answer» Correct Answer - D `y=sin(m sin^(-1)x)` Then, `(dy)/(dx)=cos(m sin^(-1)x)(m)/(sqrt(1-x^(2)))` `therefore(d^(2)y)/(dx^(2))=cos(m sin^(-1)x)m{(-1)/(2).((-2x))/((1-x^(2))^(3//2))}+(m)/(sqrt(1-x^(2))).{-sin(m sin^(-1)x)}.(m)/(sqrt(1-x^(2)))` `=(m)/(sqrt(1-x^(2)))[(x)/((1-x^(2)))cos(m sin^(-1)x)]` `-(m)/(sqrt(1-x^(2)))sin(m sin^(-1)x)]` Now, `(d^(2)y)/(dx^(2))" at "x=0" is "m[0-0]=0" "(because sin^(-1)0=0)`