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Pure · Differentiation

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Implicit differentiation

Differentiating a relation that mixes x and y — like x² + y² = 25 — without solving for y, by differentiating both sides with respect to x and treating y as a function of x (so d/dx of y² is 2y dy/dx).

Pure · Differentiation

Implicit differentiation

Differentiating a relation that mixes x and y — like x² + y² = 25 — without solving for y, by differentiating both sides with respect to x and treating y as a function of x (so d/dx of y² is 2y dy/dx).

Why it works

Not every curve can be tidied into y=f(x)y = f(x). The circle x2+y2=25x^2 + y^2 = 25 would split into two half-curves y=±25−x2y = \pm\sqrt{25 - x^2}; relations like x3+y3=6xyx^3 + y^3 = 6xy resist rearranging at all. Implicit differentiation finds the gradient without ever isolating yy.

The one idea you need: yy is still a function of xx, even if you can't see the formula. So whenever you differentiate a term containing yy, the chain rule fires and a dydx\frac{dy}{dx} drops out. Differentiating y2y^2 with respect to xx: ddx(y2)=ddy(y2)⋅dydx=2ydydx.\frac{d}{dx}(y^2) = \frac{d}{dy}(y^2)\cdot\frac{dy}{dx} = 2y\frac{dy}{dx}. The "2y2y" is the ordinary derivative; the "dydx\frac{dy}{dx}" is the chain-rule tax for yy being a function of xx. Likewise ddx(y3)=3y2dydx\frac{d}{dx}(y^3) = 3y^2\frac{dy}{dx} and ddx(sin⁡y)=cos⁡ydydx\frac{d}{dx}(\sin y) = \cos y\frac{dy}{dx}. Terms in xx alone behave normally; a plain constant differentiates to 00.

Mixed xyxy terms need the product rule. A term like xyxy is a product of two functions of xx, so ddx(xy)=1⋅y+xdydx=y+xdydx.\frac{d}{dx}(xy) = 1\cdot y + x\frac{dy}{dx} = y + x\frac{dy}{dx}.

The method is then a routine three steps:
  1. Differentiate every term of the equation with respect to xx (each yy-term
sprouts a dydx\frac{dy}{dx}).
  1. Collect all the dydx\frac{dy}{dx} terms on one side, everything else on the other.
  2. Factor out dydx\frac{dy}{dx} and divide.
The answer is usually in terms of both xx and yy — which is fine: to get a numerical gradient you substitute the coordinates of the point.

Take the circle x2+y2=25x^2 + y^2 = 25. Differentiating: 2x+2ydydx=02x + 2y\frac{dy}{dx} = 0, so dydx=−xy\frac{dy}{dx} = -\dfrac{x}{y}. At (3,4)(3, 4) that is −34-\tfrac34 — and it matches the geometry, since the tangent to a circle is perpendicular to the radius (gradient of radius 43\tfrac43, tangent −34-\tfrac34).