MA441-Lecture 14

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MA441- LEC 14 -2.6 Implicit Differentiation.

By KWANG

Goal

Implicit Differentiation.

Maple file

MAPLE FILE

Why?

Find the slope of tangent line to the circle \(x^2+y^2=25\) at the point \((3,4)\).

We can usually write \(y=f(x)\) locally in a explicit way.

\( y^2=25-x^2\) implies \(y=\pm \sqrt{25-x^2}\).

Since \(4>0\), choose \(y=\sqrt{25-x^2}\).

From the chain rule, by letting \(u=25-x^2\),
\[ \frac{dy}{dx}=\frac{dy}{dy}\cdot \frac{du}{dx}=(\frac{1}{2\sqrt{u}})\cdot(-2x)=\frac{-x}{\sqrt{25-x^2}} \]

Therefore \(y'=\frac{-1(3)}{\sqrt{25-(3)^2}}=-\frac{3}{4}\).

Problem

It takes too much time to find an explicit form from the implicit form.

There is an alternative way which is called the method of implicit differentiation.

implicit differentiation

The key assumption is \(y=f(x)\) is a function of one variable \(x\).

For simplicity, let \(y'=\frac{dy}{dx},~x'=\frac{dx}{dx}=1\).

a. Differentiate both sides of the equation \(x^2+y^2=25\) with respect to \(x\). \[ \frac{d}{dx}(x^2+y^2)=\frac{d}{dx}(25) \]

b. Use the chain rule. \[ 2x\cdot x'+2y\cdot y'=0 \]

Since \(x'=1\), \(\frac{dy}{dx}=y'=\frac{-2x}{2y}=-\frac{x}{y}\). \[slope=y'=-\frac{3}{4}\]

Ex1

a. Find \(y'\) if \(x^3+y^3=6xy\)

b. Find the tangent to the folium of Descartes \(x^3+y^3=6xy\) at the point \((3,3)\). SAGE1 SAGE2
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