Implicit Differentiation Calculator
Calculate implicit differentiation instantly with our math tool. Shows detailed work, formulas used, and multiple solution methods.
Reviewed by Manoj Kumar, Mathematics Educator
Formula
dy/dx = -(dF/dx) / (dF/dy)
For F(x,y) = ax^m*y^n + bx^p + cy^q + d = 0, differentiate both sides with respect to x using the chain rule, then solve for dy/dx. The result is the negative ratio of the partial derivative with respect to x divided by the partial derivative with respect to y.
Worked Examples
Example 1: Circle: x^2 + y^2 = 25
Problem:Find dy/dx for x^2 + y^2 = 25 at the point (3, 4).
Solution:F(x,y) = x^2 + y^2 - 25 = 0\nCoefficients: a=0 (no mixed term), b=1, p=2, c=1, q=2, d=-25\nActually: F = x^2 + y^2 - 25\ndF/dx = 2x = 2(3) = 6\ndF/dy = 2y = 2(4) = 8\ndy/dx = -(dF/dx)/(dF/dy) = -6/8 = -0.75\nTangent line: y - 4 = -0.75(x - 3), i.e., y = -0.75x + 6.25\nNormal line slope: 1/0.75 = 1.333
Result:dy/dx = -3/4 = -0.75 at (3,4) | Tangent: y = -0.75x + 6.25 | Normal slope: 4/3
Example 2: Mixed Term: x^2*y + y^2 = 4
Problem:Find dy/dx and d2y/dx2 for x^2*y + y^2 - 4 = 0 at point (1, -1+sqrt(5)).
Solution:F(x,y) = x^2*y + y^2 - 4\ndF/dx = 2xy\ndF/dy = x^2 + 2y\nAt (1, 2): dF/dx = 2(1)(2) = 4, dF/dy = 1 + 4 = 5\ndy/dx = -4/5 = -0.8\nFor d2y/dx2: use formula with Fxx=2y, Fyy=2, Fxy=2x\nd2y/dx2 = -(2y*25 - 2*2x*4*5 + 2*16)/125\n= -(100 - 80 + 32)/125 = -52/125 = -0.416
Result:dy/dx = -0.8 | d2y/dx2 = -0.416 | Concave down at this point
Frequently Asked Questions
What is implicit differentiation and when do you use it?
Implicit differentiation is a technique used to find the derivative dy/dx when y is not explicitly isolated as a function of x. Instead of having y = f(x), you have a relationship F(x,y) = 0 where y is implicitly defined. You differentiate both sides of the equation with respect to x, applying the chain rule to terms containing y (since y is a function of x), and then solve for dy/dx. This technique is essential for curves like circles (x^2 + y^2 = r^2), ellipses, hyperbolas, and any equation where isolating y is difficult or impossible. It extends naturally to higher-order derivatives and multivariable functions.
How does the chain rule apply in implicit differentiation?
The chain rule is the fundamental mechanism behind implicit differentiation. When differentiating a term containing y with respect to x, you must remember that y itself is a function of x. So when you differentiate y^n, you get n*y^(n-1)*dy/dx by the chain rule, not simply n*y^(n-1). Similarly, for a term like x^2*y^3, you apply the product rule combined with the chain rule: d/dx(x^2*y^3) = 2x*y^3 + x^2*3y^2*dy/dx. After differentiating every term in the equation, you collect all terms containing dy/dx on one side, factor out dy/dx, and solve algebraically. This systematic process works for any implicit relationship.
How do you find the second derivative using implicit differentiation?
Finding the second derivative d2y/dx2 requires differentiating dy/dx itself with respect to x, again using implicit differentiation. First, find dy/dx = -Fx/Fy where Fx and Fy are partial derivatives of F(x,y). Then differentiate this quotient using the quotient rule, remembering that both the numerator and denominator contain y, which depends on x. The formula is: d2y/dx2 = -(Fxx*Fy^2 - 2*Fxy*Fx*Fy + Fyy*Fx^2) / Fy^3. This involves second partial derivatives Fxx, Fyy, and the mixed partial Fxy. The second derivative tells you about concavity and inflection points of the implicitly defined curve.
What is the relationship between implicit differentiation and partial derivatives?
Implicit differentiation is intimately connected to partial derivatives through the implicit function theorem. For an equation F(x,y) = 0, the derivative dy/dx equals the negative ratio of partial derivatives: dy/dx = -(dF/dx)/(dF/dy), provided dF/dy is not zero. This formula is derived by differentiating F(x,y(x)) = 0 with respect to x using the multivariable chain rule: dF/dx + (dF/dy)(dy/dx) = 0. This connection extends to higher dimensions: for F(x,y,z) = 0, you can find dz/dx = -(dF/dx)/(dF/dz) and dz/dy = -(dF/dy)/(dF/dz). The implicit function theorem also specifies conditions under which such implicit functions exist and are differentiable.
References
Reviewed by Manoj Kumar, Mathematics Educator · Editorial policy