BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 3.5, Problem 16E
To determine

To calculate: The derivative of the function sinx+cosy=sinxcosy by implicit differentiation.

Expert Solution

Answer to Problem 16E

The derivative of the function is y'=cosx(cosy1)siny(sinx1) .

Explanation of Solution

Given information:

The function sinx+cosy=sinxcosy .

Formula used:

Thechain rule for differentiation is if f is a function of gthen ddx(f(g(x)))=f'(g(x))g'(x) .

Power rule for differentiation is ddxxn=nxn1 .

Product rule for differentiation is ddx(fg)=f'(x)g(x)+f(x)g'(x) where f and g are functions of x .

Calculation:

Consider the function sinx+cosy=sinxcosy .

Differentiate both sides with respect to x ,

  ddx(sinx+cosy)=ddx(sinxcosy)ddx(sinx)+ddx(cosy)=ddx(sinxcosy)

Recall that power rule for differentiation is ddxxn=nxn1 and chain rule for differentiation is if f is a function of gthen ddx(f(g(x)))=f'(g(x))g'(x) .

Also for the terms of the above expression, apply the product rule for differentiation.

Recall that product rule for differentiation is ddx(fg)=f'(x)g(x)+f(x)g'(x) where f and g are functions of x .

Apply it. Also observe that y is a function of x,

  ddx(sinx+cosy)=ddx(sinxcosy)ddx(sinx)+ddx(cosy)=ddx(sinxcosy)cosxsinyy'=cosxcosysinxsinyy'

Isolate the value of y' on left hand side and simplify,

  cosxsinyy'=cosxcosysinxsinyy'(sinxsinysiny)y'=cosxcosycosxy'=cosxcosycosxsinxsinysinyy'=cosx(cosy1)siny(sinx1)

Thus, the derivative of the function is y'=cosx(cosy1)siny(sinx1) .

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