how would i go about differentiating something like (-y sinx)(y2+1)-1/2 with respect to x? Thanks!!
M mathNewbie New member Joined Feb 10, 2012 Messages 10 Feb 10, 2012 #1 how would i go about differentiating something like (-y sinx)(y2+1)-1/2 with respect to x? Thanks!!
D Deleted member 4993 Guest Feb 10, 2012 #2 mathNewbie said: how would i go about differentiating something like (-y sinx)(y2+1)-1/2 with respect to x? Thanks!! Click to expand... d/dx[(-y sinx)(y2+1)-1/2 ] = [-y'*sinx -y*cosx]*(y2+1)-1/2 + (-y sinx)(-1/2)(y2+1)-3/2(2*y*y') Now simplify and isolate y'
mathNewbie said: how would i go about differentiating something like (-y sinx)(y2+1)-1/2 with respect to x? Thanks!! Click to expand... d/dx[(-y sinx)(y2+1)-1/2 ] = [-y'*sinx -y*cosx]*(y2+1)-1/2 + (-y sinx)(-1/2)(y2+1)-3/2(2*y*y') Now simplify and isolate y'
M mathNewbie New member Joined Feb 10, 2012 Messages 10 Feb 10, 2012 #3 Subhotosh Khan said: d/dx[(-y sinx)(y2+1)-1/2 ] = [-y'*sinx -y*cosx]*(y2+1)-1/2 + (-y sinx)(-1/2)(y2+1)-3/2(2*y*y') Now simplify and isolate y' Click to expand... why did you differentiate (-y sinx) to (-ysinx - ycosx)? isnt (-y sinx) differentiated to just (-y cos x)? thanks!
Subhotosh Khan said: d/dx[(-y sinx)(y2+1)-1/2 ] = [-y'*sinx -y*cosx]*(y2+1)-1/2 + (-y sinx)(-1/2)(y2+1)-3/2(2*y*y') Now simplify and isolate y' Click to expand... why did you differentiate (-y sinx) to (-ysinx - ycosx)? isnt (-y sinx) differentiated to just (-y cos x)? thanks!
