calc help (implicit differentiation!?

Question

Given (√x+y) - 3x=1, determine dy/dx (implicit differentiation)

Answer 1

Differentiate each term with respect to x:

d/dx (√[x + y] ) - d/dx (3x) = d/dx (1)

.5*(x + y)^-.5 * d/dx(x+y) - 3 = 0

.5*(x+y)^-.5 * dy/dx - 3 =0

dy/dx = 6/√[x + y]

Answer 2

You need to use the chain rule to differentiate the first part implicitly.
d/dx(sqrt(x + y) - 3x) = d/dx(1)
=> d/dx(sqrt(x + y) - 3 = 0
Chain rule says to take the derivative of the outside function (in this case the square root function), leaving the inside alone, and multiply it by the derivative of the inside. Thus you get:
=> [(1/2)(x + y)^(-1/2)]*(1 + dy/dx) - 3 = 0
Distributing and multiplying through by 2 yields:
=> (x + y)^(-1/2) + (x + y)^(-1/2)*dy/dx - 6 = 0
Subtract to isolate the dy/dx term:
=> (x + y)^(-1/2)*dy/dx = -(x + y)^(-1/2) + 6
Now divide both sides both sides by the coefficient of dy/dx:
=> dy/dx = [-(x + y)^(-1/2) + 6]/[(x + y)^(-1/2)]
= -1 + 6(x + y)^(1/2) = -1 + 6*sqrt(x + y)
Done. Hope this helps.

Answer 3

sqrt(x) + y - 3x = 1

x^(1/2) + y - 3x = 1

(1/2)x^(-1/2) + 1 y' - 3 = 0

y' = -(1/2)x^(-1/2) + 3

y' = -1 / [ 2sqrt(x) ] + 3

or

sqrt(x + y) - 3x = 1

(x + y)^(1/2) - 3x = 1

(1/2)(1 + y')(x + y)^(-1/2) - 3 = 0

(1/2)(1 + y')(x + y)^(-1/2) = 3

(1 + y')(x + y)^(-1/2) = 6

1 + y' = 6(x + y)^(-1/2)

y' = -1 + 6(x + y)^(1/2)

Answer 4

whilst utilising implicit differentiation, you may derive the two variables commonly, however denote dy/dx because of fact the spinoff of y each time y is derived. i will use y' to indicate dy/dx. cos(x-y)=y(2x+a million)^3 <-----Derive like usual, interior function, exterior function, etc -(a million - y')sin(x-y) = y'(2x+a million)^3 + 6y(2x+a million)^2 <------- whilst deriving y, you get a million*y'; on the stunning part, you have a function circumstances a function; now simplify -sin(x-y) + y'sin(x-y) = y'(2x+a million)^3 + 6y(2x+a million)^2 <------ Now get the y's on the comparable part y'sin(x-y) - y'(2x+a million)^3 = 6y(2x+a million)^2 + sin(x-y) <------ component out the y' y'[sin(x-y) - (2x+a million)^3] = 6y(2x+a million)^2 + sin(x-y) <------- Divide y' = [6y(2x+a million)^2 + sin(x-y)]/[sin(x-y) - (2x+a million)^3] desire I helped!