M Marcia New member Joined Oct 18, 2005 Messages 14 Oct 18, 2005 #1 Find dy/dx of y=Square root of (x^2 -2) / x^3 at x=Square root of (3). dy/dx = d/dx = x^3(1/2(x^2 - 2)^(-1/2) - 3x(Square root of (x^2 -2) Am I starting this right? What do i do now?
Find dy/dx of y=Square root of (x^2 -2) / x^3 at x=Square root of (3). dy/dx = d/dx = x^3(1/2(x^2 - 2)^(-1/2) - 3x(Square root of (x^2 -2) Am I starting this right? What do i do now?
U Unco Senior Member Joined Jul 21, 2005 Messages 1,134 Oct 18, 2005 #2 Am I starting this right? Click to expand... Yep! Provided you meant dy/dx = x^3(1/2(x^2 - 2)^(-1/2) - (3x^2)(Square root of (x^2 -2) (1)) Click to expand... Simplify x^3(1/2(x^2 - 2)^(-1/2), and then subsitute x = sqrt(3) into (1).
Am I starting this right? Click to expand... Yep! Provided you meant dy/dx = x^3(1/2(x^2 - 2)^(-1/2) - (3x^2)(Square root of (x^2 -2) (1)) Click to expand... Simplify x^3(1/2(x^2 - 2)^(-1/2), and then subsitute x = sqrt(3) into (1).
