implicit differentiation question?

The idea is that you cannot always separate the y from the x. So you assume y is a function of x, like you would use u in substitution problems. Ex d(y^2)=2y*dy/dx by chain rule

So if y is really sin x , then d(sin x)^2=2sin x*d(sin x)

In this problem, find the derivative term by term

D(x^2)=2x

D(xy)=(product rule)=x*dy/dx+y*1

d(2x)=2

d(3x^2)=6x

D(4)=0

so the equation becomes

2x+x*dy/dx+y+2+6x=0

now treat dy/dx as the variable, since that is what you are trying to find.

X*dy/dx=-8x-y-2

Dy/dx=(-8x-y-2)/x

one thing that confuses people is that the derivative of any variable is the prime of that variable, not necessarily one. since in text books, almost all functions are in terms of x and being differentiated with respect to x, the derivative of it with respect to itself is one. this makes sense if you think of the quotient itself which is dx/dx. when you differentiate y with respect to itself, it is dy/dx, which is not equal to 1, so we use y' for the sake of simplicity.

for example, given the function y=x^2, you would probably look and say y'=2x, which is correct, but you arent including all the steps which is: dy/dx=2x*dx/dx. technically, x is an inside function, and you need to use the chain rule, but since it is often just 1 (whenever the inside function is x), we just say in our head that we arent using the chain rule.

if it makes it easier, during implicit differentiation, just write out all of the variables, or even the differentials (dx/dx for example).

x^2+xy+2x+3x^2=4

2x*dx/dx +dx/dx*y+x*dy/dx +2*dx/dx +6x*dx/dx =0

2x*1 +1*y+x*y' +2*1 +6x*1 =0

2x+y+xy'+2+6x=0

xy'+y+8x+2=0

since you started with an implicit equation, you can generally leave the answer without solving for y. on a test though, i probably would just to be safe, or if you needed to graph the line using a calculator. here:

y'=(-y-8x-2)/x

and solving the starting equation for y and substituting it into the derivative equation gives

y=(4-4x^2-2x)/x

y'=(-(4-4x^2-2x)/x-8x-2)/x

Where is the question. You have given an equation but not mentioned what you want! Find dy/dx or dx/dy or something else!

Differentiate the function implicitly w/respect to x. Make sure to use chain and product rules!

2x + x(dy/dx) + y + 2 + 6x = 0

x(dy/dx) + 8x + y + 2 = 0

x(dy/dx) = -8x - y - 2

dy/dx = (-8x - y - 2)/x

I hope this helps!

only differentiate the finished subject i think of the equation your watching is; 5x^3 + x^2*y - x * y^3 = 2 Differentiate to get 15 * x^2 * dx + 2 * x * y * dx + x^2 * dy - y^3 * dx - 3 * x * y^2 * dy = 0 divide via dx and assemble like words (x^2 - 3xy^2) * (dy/dx) = (y^3 -15x^2 -2xy) dy/dx = (y^3 -15x^2 -2xy)/(x^2 - 3xy^2) and your achieved