Question

nth derivative of sin inverse x

Answer 1

1st derivative:\[{1\over \sqrt{1-x^2}}=\left( 1-x^2 \right)^{-\frac12}\]

2st derivative : \[-\frac12 \times \left( 1-x^2 \right)^{-\frac32} \times-2x = x \left( 1-x^2 \right)^{-\frac32}\]

3rd derivative : \[\left( 1-x^2 \right)^{-3/2}+x \cdot-\frac32\cdot \left( 1-x^2 \right)^{-5/2}\cdot -2x\\ = {1-x^2+3x^2 \over \left( 1-x^2 \right)^{5/2}}\\= {1+2x^2 \over \left( 1-x^2 \right)^{5/2}}\]