# The derivative of the function 1 + x = sin ( x y 2 ) by implicit differentiation.

### Single Variable Calculus: Concepts...

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

### Single Variable Calculus: Concepts...

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

#### Solutions

Chapter 3.5, Problem 10E
To determine

## To calculate: The derivative of the function 1+x=sin(xy2) by implicit differentiation.

Expert Solution

The derivative of the function is 1y2cos(xy2)2xycos(xy2) .

### Explanation of Solution

Given information:

The function 1+x=sin(xy2) .

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 1+x=sin(xy2) .

Differentiate both sides with respect to x ,

ddx(1+x)=ddx(sin(xy2))ddx(1)+ddx(x)=ddx(sin(xy2))

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(1+x)=ddx(sin(xy2))ddx(1)+ddx(x)=ddx(sin(xy2))0+1=cos(xy2)[y2+2xyy']1=y2cos(xy2)+2xycos(xy2)y'

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

y2cos(xy2)+2xycos(xy2)y'=12xycos(xy2)y'=1y2cos(xy2)y'=1y2cos(xy2)2xycos(xy2)

Thus, the derivative of the function is 1y2cos(xy2)2xycos(xy2) .

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