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3.23 Gradient of the tangent to a curve

We can find $dy/dx$ in terms of $f$ without solving for $y$ .

Proposition.

Let $f(x,y)=0$ define a curve $C$ such that $y$ is implicitly a differentiable function of $x$ . Then the gradient of the tangent to $C$ at $(x,y)$ on $C$ is

{{dy}\over{dx}}=-{{f_{x}}\over{f_{y}}}.

Note that there may be several values for $dy/dx$ at any given $x$ , since there may be several choices for $y$ on the curve above $x$ .