Let's do this step-by-stepHow do you solve this?
“If (x+y)sin(xy)=1, use implicit differentiation to determine dy/dx, expressing your answer in the form dy/dx = f(x,y), where f(x,y) is an expression that involves terms in both x and y.”
dx/dy appears twice how do I solve for dy/dx = f(x,y)? ConfusedLet's do this step-by-step
d/dx (x*y) = x * dy/dx + y ..............(1)............ product rule
d/dx[sin(xy)] = [cos(xy)] * d/dx (x*y) = [x * dy/dx + y] * cos(x*y) ..........................(2) ..........................chain-rule
d/dx (x + y) = 1 + dy/dx.....................................(3)
d/dx [(x+y) * sin(x*y)] = [d/dx (x+y)] * sin(xy) + (x + y) * d/dx [sin(xy)] ..............................................(4) ............... product rule
Now put (1), (2) and (3) into (4)
You tell us.What is dy/dx when solved?
One thing that sometimes helps is to replace dy/dx with some variable, maybe A to make it stand out:What is dy/dx when solved?
1) How does (y + x*dy/dx) * sin(x+y) become y*sin(x+y) + x*sin(x+y)*dy/dxTrue I am working on my algebra, these problems really highlights any gaps in my knowledge (lots). I understand how the chain rule and product rule has been applied. I don’t understand how dy/dx (or A) was transposed/manipulated, what algebraic rules were used to do this? I researched over the weekend but can’t find any other examples showing a derivative being isolated this way.
1) How does (y+xdy/dx)sin(x+y) become ysin(x+y)+xsin(x+y)dy/dx?
2) How does xycos(x+y)(1+dy/dx) become
xycos(x+y)+xycos(x+y)dy/dx?
Yes. A derivative is just a function.maybe you write dy/dx as y'
in the above you just distribute into the parenthesis