How to do d/dx cos(xy)?
I think you're supposed to find the derivative... I know d/dx (xy) = y + xy', but what is d/dx cos(xy)? Sorry for the confusion!
BQ: What would be d/dx √(xy)?
Update:EDIT: It's called implicit differentiation, oops!
I think you're supposed to find the derivative... I know d/dx (xy) = y + xy', but what is d/dx cos(xy)? Sorry for the confusion!
BQ: What would be d/dx √(xy)?
Update:EDIT: It's called implicit differentiation, oops!
Comments
By the chain rule: dcos(u)/dx = -sin(u)du/dx
If u = xy, by the product rule, du/dx = ydx/dx + xdy/dx = y + xdy/dx
dcos(xy)/dx = -sin(xy)(y + xdy/dx) = -(y+xy')sin(xy)
By the same process, chain rule combined with product rule:
d/dx √u = [1/(2√u)] du/dx
d/dx √(xy) = (y+xy')/[2√(xy)]