3.2 Implicit Differentiation

Syllabus
2020
Topic
3.2
Level

Differentiate an Implicit Relation

An implicit equation links xx and yy without necessarily isolating yy. Treat yy as a differentiable function y(x)y(x). When a term involving yy is differentiated with respect to xx, the chain rule introduces a factor of dy/dxdy/dx.

\frac{d}{dx}[F(y)]=F'(y)\frac{dy}{dx}

  1. Differentiate every term on both sides with respect to xx.
  2. Attach dy/dxdy/dx whenever the chain rule differentiates a function of yy.
  3. Move all terms containing dy/dxdy/dx to one side.
  4. Factor out dy/dxdy/dx and solve for it.

For x2+y2=25x^2+y^2=25, differentiate with respect to xx: 2x+2ydydx=0.2x+2y\frac{dy}{dx}=0. Therefore 2ydydx=2xdydx=xy.2y\frac{dy}{dx}=-2x\quad\Rightarrow\quad\frac{dy}{dx}=-\frac{x}{y}. At (3,4)(3,4), the slope is 3/4-3/4.

Do not differentiate y2y^2 as merely 2y2y: because yy depends on xx, the result is 2ydy/dx2y\,dy/dx. The solved formula x/y-x/y applies where y0y\ne 0; when y=0y=0 on this circle, the tangent is vertical rather than having a finite dy/dxdy/dx.