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 36E
To determine

Find the derivative of the given function.

Expert Solution

Answer to Problem 36E

The derivative of the given function is dydx=2xln(2)ln(3)23x23x2 .

Explanation of Solution

Given:

The given function is y=23x2 .

Calculation:

  y=23x2

Use exponent rule xy=eyln(x) .

  23x2=e3x2ln(2)y=e3x2ln(2)

Apply chain rule.

  df(a)dx=dfdadadx

Let f=ea,a=3x2ln(2)

  dydx=dda(ea)ddx(3x2ln(2))

Use derivative rule.

  ddx(ea)=ea .

  dydx=eaddx(3x2ln(2))

Substitute the value of a=3x2ln(2) .

  dydx=e3x2ln(2)ddx(3x2ln(2))dydx=e3x2ln(2)ln(2)[ddx(3x2)]

Use exponent rule xy=eyln(x) .

  3x2=ex2ln(3)

Apply chain rule.

  df(a)dx=dfdadadx

Let f=ea,a=x2ln(3)

  dydx=e3x2ln(2)ln(2)[dda(ea)ddx(x2ln(3))]

Use derivative rule.

  ddx(ea)=ea and ddx(xn)=nxn1

  dydx=e3x2ln(2)ln(2)[ea2xln(3)]

Substitute the value of a=x2ln(3) .

  dydx=e3x2ln(2)ln(2)[ex2ln(3)2xln(3)]dydx=2xln(2)ln(3)e3x2ln(2)ex2ln(3)dydx=2xln(2)ln(3)23x23x2

Hence the derivativeof the given function is dydx=2xln(2)ln(3)23x23x2 .

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