   Chapter 3.5, Problem 25E ### 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.25. y sin 2x = x cos 2y, (π/2, π/4)

To determine

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

Explanation

Given:

The curve is ysin2x=xcos2y.

The point is (π2,π4).

Derivative rules:

Chain rule

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

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:

Consider ysin2x=xcos2y.

Differentiate the above equation implicitly with respect to x,

yddx(sin2x)+sin2xddx(y)=xddx(cos2y)+cos2yddx(x)yddx(sin2x)+sin2xdydx=xddx(cos2y)+cos2y(1)

Apply the chain rule (1) and simplify the terms,

y[dd(2x)(sin2x)ddx(2x)]+sin2xdydx=x[dd(2y)(cos2y)ddx(2y)]+cos2yy[cos2x(2)]+sin2xdydx=x[2sin2ydydx]+cos2y2ycos2x+sin2xdydx=2xsin2ydydx+cos2y

Combine the terms of dydx,

sin2xdydx+2xsin2ydy

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Study Guide for Stewart's Single Variable Calculus: Early Transcendentals, 8th 