Implicit Differentiation

Suppose the equation $u(x,y)=0$ defines the relation between variables $x$ and $y$. Thinking of $x$ as the independent variable, we apply the chain rule to differentiate the equation and then solve the resulting equation for the derivative $dy/dx$. This process is called implicit differentiation. The process results in the equation

Exercise: Find the equation of the tangent line to the graph of the equation at the given points:

(a)$x^2+y^2=25;(3,-4)$ (b) $xy=-8;(4,-2)$

(c) $x^2y=x+2;(2,1)$ (d) MATH

(e) $xy^2+x^2y=2;(1,-2)$ (f) MATH

(g) MATH (h) MATH

(i) MATH (j) MATH

Higher-Order Derivatives of an Implicit Function

If $y=y(x)$ satisfies the equation $u(x,y)=0$, then MATH. This is the result of the chain rule MATH. To find the second derivative MATH, apply the formula

i.e., replace $u$ and $\frac{dy}{dx}$ in MATHbyMATH . To find the third derivative MATH, replace $u$ in MATH by MATH and $\frac{dy}{dx}$ by $y^{\prime }(x).$