# The derivative of the function e y cos x = 1 + sin ( x y ) 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 15E
To determine

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

Expert Solution

The derivative of the function is y'=ycos(xy)+eysinxeycosxxcos(xy) .

### Explanation of Solution

Given information:

The function eycosx=1+sin(xy) .

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 eycosx=1+sin(xy) .

Differentiate both sides with respect to x ,

ddx(eycosx)=ddx(1+sin(xy))

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(eycosx)=ddx(1+sin(xy))eycosxy'eysinx=cos(xy)[y+xy']eycosxy'eysinx=ycos(xy)+xcos(xy)y'

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

eycosxy'eysinx=ycos(xy)+xcos(xy)y'(eycosxxcos(xy))y'=ycos(xy)+eysinxy'=ycos(xy)+eysinxeycosxxcos(xy)

Thus, the derivative of the function is y'=ycos(xy)+eysinxeycosxxcos(xy) .

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