y2=f(x) then evaluate dy/dx(y3.d2y/dx2)

Question

it is a three degree polynomial

Answer 1

y²=f(x), y=√f(x).'

' denotes d/dx, and " denotes d²/dx².

2yy'=f'(x), y'=f'(x)/2y.

2y'²+2yy"=f"(x),

y"=(f"(x)-2y'²)/2y=(f"(x)-2(f'(x)/2y)²)/2y.

y'y³y"=(f'(x)/2y)y³(f"(x)-2(f'(x)/2y)²)/2y,

y'y³y"=(f'(x)/2y)y³(2y²f"(x)-(f'(x))²)/(4y³),

y'y³y"=(f'(x)/(8√f(x))(2f(x)f"(x)-(f'(x))²),

y'y³y"=f'(x)√f(x)(2f(x)f"(x)-(f'(x))²)/(8f(x)).