   Chapter 3.5, Problem 37E ### 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

# Find y″ by implicit differentiation.37. sin y + cos x =1

To determine

To find: The derivative y by implicit differentiation.

Explanation

Given:

The equation siny+cosx=1.

Derivative rules:

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

(3) Quotient Rule: If f1(x) and f2(x) are both differentiable, then ddx[f1(x)f2(x)]=f2(x)ddx[f1(x)]f1(x)ddx[f2(x)][f2x]2

Calculation:

Obtain the first derivative of the equation.

Consider the equation siny+cosx=1.

Differentiate the equation implicitly with respect to x,

ddx(siny+cosx)=ddx(1)ddx(siny)+ddx(cosx)=0ddx(siny)sinx=0

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

[ddy(siny)dydx]sinx=0cosydydxsinx=0cosydydx=sinxdydx=sinxcosy

Therefore, the derivative of the equation is dydx=sinxcosy.

Obtain the second derivative of the equation.

The first derivative is y(x)=sinxcosy

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