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 12E
To determine

To calculate: The derivative of the function ysin(x2)=xsin(y2) by implicit differentiation.

Expert Solution

Answer to Problem 12E

The derivative of the function is sin(y2)2xycos(x2)sin(x2)2xycos(y2) .

Explanation of Solution

Given information:

The function ysin(x2)=xsin(y2) .

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 ysin(x2)=xsin(y2) .

Differentiate both sides with respect to x ,

  ddx(ysin(x2))=ddx(xsin(y2))

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(ysin(x2))=ddx(xsin(y2))sin(x2)y'+ycos(x2)2x=sin(y2)+xcos(y2)2yy'

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

  [sin(x2)xcos(y2)2y]y'=sin(y2)ycos(x2)2xy'=sin(y2)ycos(x2)2xsin(x2)xcos(y2)2yy'=sin(y2)2xycos(x2)sin(x2)2xycos(y2)

Thus, the derivative of the function is sin(y2)2xycos(x2)sin(x2)2xycos(y2) .

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