# 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 31E
To determine

## Find the derivative of the given function.

Expert Solution

The derivative of the given function is dydx=π2sin(πx)ln(2)cos(πx) .

### Explanation of Solution

Given:

The given function is y=2sin(πx) .

Calculation:

Use exponent rule xy=eyln(x) .

y=2sin(πx)2sin(πx)=esin(πx)ln(2)y=esin(πx)ln(2)

Apply chain rule.

Let f=ea,a=sin(πx)ln(2)

dydx=dda(ea)ddx[sin(πx)ln(2)]

Use derivative rule.

ddx(ex)=exandddx(sinmx)=mcosmx .

dydx=ea[πcos(πx)ln(2)]dydx=πealn(2)cos(πx)

Substitute the value of a=sin(πx)ln(2) .

dydx=πesin(πx)ln(2)ln(2)cos(πx)dydx=π2sin(πx)ln(2)cos(πx)

Hence the derivativeof the given function is dydx=π2sin(πx)ln(2)cos(πx) .

### Have a homework question?

Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers!