How do you find #(d^2y)/(dx^2)# for #2=2x^2-4y^2#?
1 Answer
Feb 28, 2017
Explanation:
differentiate all terms on both sides
#color(blue)"implicitly with respect to x"#
#rArr0=4x-8y.dy/dx#
#rArrdy/dx=(-4x)/(-8y)=x/(2y)# The second derivative is obtained by differentiating
#dy/dx# differentiate
#dy/dx" using the " color(blue)"quotient rule"#
#rArr(d^2y)/(dx^2)=(2y.1-x.2dy/dx)/(4y^2)#
#color(white)(rArr(d^y)/(dx^2))=(2y-2x(x/(2y)))/(4y^2)#
#color(white)(rArr(d^2y)/(dx^2))=(2y-((x^2)/y))/(4y^2)#
#color(white)(rArr(d^2y)/(dx^2))=(2y^2-x^2)/(4y^3)#