BuyFind

Single Variable Calculus: Concepts...

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

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

Answer to Problem 31E

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.

  df(a)dx=dfdadadx

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) .

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