# Find the derivative of the given function.

### 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 3.4, Problem 35E
To determine

## Find the derivative of the given function.

Expert Solution

The derivative of the given function is dydx=πsin(sin(tanπx))cos(tanπx)sec2πx2sin(tanπx) .

### Explanation of Solution

Given:

The given function is y=cossin(tanπx) .

Calculation:

y=cossin(tanπx)

Apply chain rule.

Let f=cosa,a=sin(tanπx)

dydx=dda(cosa)ddx(sin(tanπx))

Use derivative rule.

ddx(cosx)=sinx .

dydx=sin(a)ddx(sin(tanπx))

Substitute the value of a=sin(tanπx) .

dydx=sin(sin(tanπx))ddx(sin(tanπx))

Apply chain rule.

Let f=a12,a=sin(tanπx)

dydx=sin(sin(tanπx))dda(a12)ddx(sin(tanπx))

Use derivative rule.

ddx(xn)=nxn1

Substitute the value of a=sin(tanπx) .

dydx=sin(sin(tanπx))12sin(tanπx)ddx(sin(tanπx))

Apply chain rule.

Let f=sina,a=tanπx

dydx=sin(sin(tanπx))2sin(tanπx)dda(sina)ddx(tanπx)

Use derivative rule.

ddx(sinx)=cosx and ddx(tankx)=ksec2kx

dydx=sin(sin(tanπx))2sin(tanπx)cosaπsec2πx

Substitute the value of a=tanπx .

dydx=sin(sin(tanπx))2sin(tanπx)cos(tanπx)πsec2πxdydx=πsin(sin(tanπx))cos(tanπx)sec2πx2sin(tanπx)

Hence the derivativeof the given function is dydx=πsin(sin(tanπx))cos(tanπx)sec2πx2sin(tanπx) .

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