Derivatives: rules and functions

Implicit differentiation

Implicit differentiation treats yy as a function of xx, differentiates both sides of the equation with respect to xx, and solves for y′y'.

ddxf(y)=f′(y) dydx\begin{gathered} \frac{d}{dx}f(y) \\[4pt] =f'(y)\,\frac{dy}{dx} \end{gathered}

Because yy is a function of xx, differentiating a term in yy with respect to xx brings out a factor y′y' by the chain rule. For example, ddxy2=2y y′\frac{d}{dx}y^2=2y\,y'.

x2+2y2=4⇒ 2x+4y y′=0 ⇒ y′=−x2y(y≠0)\begin{gathered} x^2+2y^2=4 \\[4pt] \Rightarrow\ 2x+4y\,y'=0\ \Rightarrow\ y'=-\frac{x}{2y}\quad(y\neq 0) \end{gathered}

For a product such as xy=2xy=2, use the product rule: y+x y′=0y+x\,y'=0, so y′=−yxy'=-\frac{y}{x}.

Examples

  1. 01What is y'?x2+2y2=4\displaystyle x^2+2y^2=4AnswerHide
    Answer−x2y\displaystyle -\frac{x}{2y}
  2. 02What is y'?x2+y2=4\displaystyle x^2+y^2=4AnswerHide
    Answer−xy\displaystyle -\frac{x}{y}
  3. 03What is y'?x2+y2=1\displaystyle x^2+y^2=1AnswerHide
    Answer−xy\displaystyle -\frac{x}{y}
  4. 04What is y'?x2+3y2=6\displaystyle x^2+3y^2=6AnswerHide
    Answer−x3y\displaystyle -\frac{x}{3y}
  5. 05What is y'?xy=2\displaystyle xy=2AnswerHide
    Answer−yx\displaystyle -\frac{y}{x}
  6. 06What is y'?xy=4\displaystyle xy=4AnswerHide
    Answer−yx\displaystyle -\frac{y}{x}

FAQ

Q1Can yy appear in the answer?AnswerHide
A
Yes. With implicit differentiation, y′y' is normally written in terms of xx and yy. Substitute a point (x, y)(x,\ y) on the curve to get the slope there.
Q2Is the derivative of y2y^2 just 2y2y?AnswerHide
A
Not when differentiating with respect to xx: it is 2y y′2y\,y'. Without the y′y', you cannot solve for y′y'.

Published 2026-10-14 · Updated 2026-10-14