# The derivative of the function tan ( x − y ) = y 1 + x 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 14E
To determine

## To calculate: The derivative of the function tan(x−y)=y1+x2 by implicit differentiation.

Expert Solution

The derivative of the function is 2xy+(1+x2)2sec2(xy)1+x2+(1+x2)2sec2(xy) .

### Explanation of Solution

Given information:

The function tan(xy)=y1+x2 .

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 .

Quotient rule for differentiation is ddx(fg)=g(x)f'(x)f(x)g'(x)[g(x)]2 where f and g are functions of x .

Derivative of tanx is sec2x .

Calculation:

Consider the function tan(xy)=y1+x2 .

Differentiate both sides with respect to x ,

ddx(tan(xy))=ddx(y1+x2)

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 quotient rule for differentiation is ddx(fg)=g(x)f'(x)f(x)g'(x)[g(x)]2 where f and g are functions of x .

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

ddx(tan(xy))=ddx(y1+x2)sec2(xy)[1y']=(1+x2)y'y(2x)(1+x2)2(1+x2)2sec2(xy)(1+x2)2sec2(xy)y'=y'+x2y'2xy

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

(1+x2)2sec2(xy)(1+x2)2sec2(xy)y'=y'+x2y'2xy(x21(1+x2)2sec2(xy))y'=2xy(1+x2)2sec2(xy)y'=2xy(1+x2)2sec2(xy)1x2(1+x2)2sec2(xy)y'=2xy+(1+x2)2sec2(xy)1+x2+(1+x2)2sec2(xy)

Thus, the derivative of the function is 2xy+(1+x2)2sec2(xy)1+x2+(1+x2)2sec2(xy) .

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