Exercise 1797Calculus › Multivariable calculus › Partial derivativesL1Added 3 Nov 2026

Implicit differentiation

#implicit differentiation#Implicit Function Theorem

Statement

The curve $x^2+y^2=1$ defines $y$ implicitly as a function of $x$ near $(1/2,\sqrt3/2)$ . Compute $y'(x)$ by implicit differentiation.

Hint

Differentiate $x^2+y(x)^2=1$ with respect to $x$ .

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Differentiation. $2x+2y(x)y'(x)=0$ .

Solve. $y'(x)=-x/y(x)$ .

At $x=1/2$ . $y=\sqrt3/2$ , so $y'(1/2)=-\frac{1/2}{\sqrt3/2}=-1/\sqrt3$ .

Conclusion. $y'(1/2)=-1/\sqrt3$ , consistent with the geometric interpretation of a tangent to the circle.