y = 3e^-x
y' = -3e^-x
e^x y =3
find dy/dx using implicit differentiation
the answer is -y
Compare your answer with the result obtained by first solving for y as a function of x and then taking the derivative.
y=?
dy/dx=?
help me out with these last two please
I don't know what they are asking
6 answers
done implicitly:
e^x y = 3
e^x (dy/dx) + y e^x = 0
dy/dx = -y e^x/e^x = -y
solving for y first:
y = 3/e^x = 3 e^-x
dy/dx = 3(e^-x)(-1)
or
= -3/e^x
e^x y = 3
e^x (dy/dx) + y e^x = 0
dy/dx = -y e^x/e^x = -y
solving for y first:
y = 3/e^x = 3 e^-x
dy/dx = 3(e^-x)(-1)
or
= -3/e^x
thank you that one was correct, i just had one other i couldn't get, the problem is find dy/dx using implicit differentiation: 4x+3y=xy
i found the derivative to be...
y-4/3-x
same question as the last one,
y=?
dy/dx=?
Thanks again for your help
i found the derivative to be...
y-4/3-x
same question as the last one,
y=?
dy/dx=?
Thanks again for your help
i submitted the answer steve had first and it came out correct for the previous question
4x+3y=xy
4 + 3dy/dx = x(dy/dx) + y
3dy/dx - xdy/dx = y - 4
dy/dx(3-x) = y-4
dy/dx = (y-4)/(3-x) --- > you had that, if you had used the proper bracketing
solving the original for y ....
4x + 3y = xy
3y - xy = -4x
y(3-x) = -4x
y = -4x/(3-x) or 4x/(x-3)
now use quotient rule:
dy/dx = [4(x-3) - 4x(1) ]/(x-3)^2
= -12/(x-3)^2
4 + 3dy/dx = x(dy/dx) + y
3dy/dx - xdy/dx = y - 4
dy/dx(3-x) = y-4
dy/dx = (y-4)/(3-x) --- > you had that, if you had used the proper bracketing
solving the original for y ....
4x + 3y = xy
3y - xy = -4x
y(3-x) = -4x
y = -4x/(3-x) or 4x/(x-3)
now use quotient rule:
dy/dx = [4(x-3) - 4x(1) ]/(x-3)^2
= -12/(x-3)^2
Very nice, thanks Reiny