text: calculus problem using second derivative? if x^2 - 2y^2 = 2, then (d^2 y)/(d x^2) can be expressed in the form a y^b, where a and b are rational numbers in simplest form.  find a and b. we start by using implicit differentiation:  x² - 2y² = 2 2x - 4y dy/dx = 0 dy/dx = x/(2y)  differentiating a second time:  d²y/dx² = (2y - 2 dy/dx x)/(4y²)  substituting the value we already have for dy/dx:  d²y/dx² = (2y - 2 x/(2y) x)/(4y²) d²y/dx² = (2y - x²/y)/(4y²)  solving the original equation for x²:  x² - 2y² = 2 x²=2+2y²  and substituting:  d²y/dx² = (2y - (2+2y²)/y)/(4y²) d²y/dx² = (2y - 2/y - 2y)/(4y²) d²y/dx² =  (-2/y)/(4y²) d²y/dx² =  -1/(2y³)  d²y/dx² = -1/2 * y^(-3)  so a=-1/2 and b=-3