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Last updated on September 26, 2025
An implicit function describes the relationship between independent variables and a dependent variable. In a single equation, like y - 3x2 + 2x + 5 = 0, an explicit function shows the output clearly, but an implicit function doesn’t. Like y = 3x + 2. Implicit differentiation uses the chain rule or product rule to find the derivative without solving for y.
An implicit function includes the dependent variable within an equation, like x2+y2=1, without solving for it. Instead of giving y explicitly as y = f(x). Implicit differentiation lets you find dy/dx by differentiating both sides using the chain rule without solving for y first. Implicit differentiation is useful when it’s hard or impossible to solve for y directly.
Cannot be written as y = f(x)
The dependent variable y isn't isolated; both x and y appear together in the same equation
Always in the form f(x, y) = 0
The equation combines x and y like x2+y2-1 = 0 for a circle
Frequently non-linear and multivariable
Many implicit functions are polynomials or complex relations, including many variables
Combines dependent and independent terms.
These cannot be easily solved for y because it stays within the equation with x.
It may not pass the vertical line test.
An implicit relation does not always pass the vertical line test because:
A vertical line might cross its graph more than once.
That means y is not a single-valued function of x.
While solving an equation involving both x and y together f(x, y) = 0, you can't isolate y very easily, you directly differentiate both sides without solving a variable first. With respect to x treating y as a function of x and use the chain rule whenever you differentiate a term including y.
Differentiate each term in the equation with respect to x, adding dydx while differentiating anything with y
Collect all dydx, terms on one side
Solve for dydx, the result will usually involve both x and y
x2+xy+y=0
Differentiate both sides w.r.t.x:
ddx(x2)+ddx(xy)+ddx(y)=0
Becomes
2x+(xdydx+y1)+dydx=0
Combine dydx terms and isolate
(x+1)dydx+(2x+y)=0dydx=-2x+yx+1
You get dydx directly without needing to solve for y first.
The implicit function theorem helps to turn complicated relationships into simpler functions of real variables. It shows that, even if a relation doesn’t define a function globally, it can behave like a function locally. For example, given an equation like F(x, y)=0, the theorem ensures that we can solve for y as a smooth function of x. Certain partial derivatives are non-zero, indicating how the function changes as x changes.
Real life application is important in many fields such as economics, physics, chemistry, The uses are explained below;
While solving implicit function, students often forget that y depends on x in implicit differentiation.
Differentiate the equation x2+y2=2.
dydx=-xy
Here, we are given a relation involving both x and y. Since y is implicit of a function x, we apply the chain rule while differentiating y2. The derivative of the x2 is 2x, and the derivative of y2 is 2ydydx. After the differentiation, we only take dydx to get the final answer.
ddx(x2+y2)=ddx(2)
2x+2ydydx=0
dydx=-xy
Differentiate the equation x2+y3=4
dydx=-2x3y2
Differentiate both the sides of the equation with respect to x. Apply the power rule to x2 and the chain rule to y3(since y is a function of x). After differentiating, we rearrange the equation to isolate dydx.
ddx(x2+y3)=ddx(4)
2x+3y2dydx=0
Then,
dydx=-2x3y2
Find the derivative dydx for the equation x . y3=1
dydx=-y3x
Here, we use the product rule to differentiate x . y3. The derivative of x is said to be 1 and that of y3 is 3y2dydx. Now, simplify to separate dydx. Finally, reduce the expression.
ddx(xy3)=ddx(1)
y3+x . 3y2dydx=0
3xy2dydx=-y3
dydx=-y3x
Differentiate the equation 2y3+4x2-y=x6
dydx=6x5-8x6y2-1
Here, differentiate each term of the equation. Use the chain rule for y3 and y, and the power rule for the x2 and x6. Now, group the terms containing dydx, factor it out, and solve for it.
Here,
ddx(2y3+4x2-y)=ddx(x6)
6y2dydx+8x-dydx=6x5
(6y2-1)dydx=6x5-8x
dydx=6x5-8x6y2-1
Differentiate x2+xy+y2=4
dydx=-2x+yx+2y
Here, to differentiate xy we need the product rule, which is derivative of x is 1, and derivative of y is dydx. After applying the derivative to each term, combine the like terms and isolate dydx.
Here,
ddx(x2+xy+y2)=ddx(4)
2x+(xdydx+y)+2ydydx=0
(x+2y)dydx=-2x-y
dydx=-2x-yx+2y