Finda new identity using 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 3.3, Problem 47E

a.

To determine

Finda new identity using differentiation.

Expert Solution

The new identity is ddx(tanx)=sec2x .

Explanation of Solution

Given:

The given relation is tanx=sinxcosx .

Calculation:

tanx=sinxcosxddx(tanx)=ddx(sinxcosx)

From R.H.S

ddx(sinxcosx)

Use quotient rule.

(fg)'=gf'fg'g2

=cosxddx(sinx)sinxddx(cosx)cos2x

Use derivative rule.

ddx(sinx)=cosx and ddx(cosx)=sinx .

=cosxcosxsinx(sinx)cos2x=cos2x+sin2xcos2x=1cos2x=sec2x

Hence the new identityis ddx(tanx)=sec2x .

b.

To determine

Find a new identity using differentiation.

Expert Solution

The new identity is ddx(secx)=secxtanx .

Explanation of Solution

Given:

The given relation is secx=1cosx .

Calculation:

secx=1cosxddx(secx)=ddx(1cosx)

From R.H.S

ddx(1cosx)

Use quotient rule.

(fg)'=gf'fg'g2

=cosxddx(1)1ddx(cosx)cos2x

Use derivative rule.

ddx(constant)=0 and ddx(cosx)=sinx .

=cosx01(sinx)cos2x=sinxcos2x=secxtanx

Hence the new identity is ddx(secx)=secxtanx .

c.

To determine

Find a new identity using differentiation.

Expert Solution

The new identity is ddx(sinx+cosx)=cotx1cscx .

Explanation of Solution

Given:

The given relation is sinx+cosx=1+cotxcscx .

Calculation:

sinx+cosx=1+cotxcscxddx(sinx+cosx)=ddx(1+cotxcscx)

From R.H.S

ddx(1+cotxcscx)

Use quotient rule.

(fg)'=gf'fg'g2

=cscxddx(1+cotx)(1+cotx)ddx(cscx)csc2x

Use derivative rule.

ddx(constant)=0 , ddx(cotx)=csc2x and ddx(cscx)=cscxcotx .

=cscx(0csc2x)(1+cotx)(cscxcotx)csc2x=csc3x+cscxcotx+cscxcot2xcsc2x=csc2x+cotx+cot2xcscx=csc2x+cotx+csc2x1cscx=cotx1cscx

Hence the new identity is ddx(sinx+cosx)=cotx1cscx .

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