   Chapter 3.5, Problem 32E ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343

#### Solutions

Chapter
Section ### Single Variable Calculus: Early Tr...

8th Edition
James Stewart
ISBN: 9781305270343
Textbook Problem

# Use implicit differentiation to find an equation of the tangent line to the curve at the given point.32. y2(y2 – 4) = x2(x2 –5), (0. –2), (devil’s curve) To determine

To find: The equation of the tangent line to the given equation at the point.

Explanation

Given:

The curve is y2(y24)=x2(x25).

The point is (0,2).

Derivative rules:

(1) Chain rule: If y=f(u) and u=g(x)  are both differentiable function, then dydx=dydududx.

(2) Product rule: ddx(f(x)g(x))=f(x)ddx(g(x))+g(x)ddx(f(x))

Formula used:

The equation of the tangent line at (x1,y1) is, yy1=m(xx1) (1)

Where, m is the slope of the tangent line at (x1,y1) and m=dydx|x=x1,y=y1.

Calculation:

Obtain the equation of the tangent line to the equation at the point.

Rewrite the given equation as follows,

y2(y24)=x2(x25)y44y2=x45x2

Differentiate the equation implicitly with respect to x,

ddx(y44y2)=ddx(x45x2)ddx(y4)4ddx(y2)=ddx(x4)5ddx(x2)ddx(y4)4ddx(y2)=4x210x<

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