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, Problem 2RCC

(a)

To determine

To State: The derivative of y=xn.

Expert Solution

Answer to Problem 2RCC

The derivative of y=xn is, y=nxn1.

Explanation of Solution

Binomial expansion:

(x+h)n=xn+nxn1h+n(n1)2xn2h2+...+nxhn1+hn

Calculation:

Obtain the derivative of y.

y=limh0(x+h)nxnh=limh0(xn+nxn1h+n(n1)2xn2h2+...+nxhn1+hn)xnh=limh0nxn1h+n(n1)2xn2h2+...+nxhn1+hnh=nxn1

Therefore, the derivative of y is y=nxn1.

(b)

To determine

To find: The derivative of y=ex.

Expert Solution

Answer to Problem 2RCC

The derivative of y=ex is, y=ex.

Explanation of Solution

Calculation:

Obtain the derivative of y.

y=limh0f(x+h)f(x)h=limh0ex+hexh=limh0ex(eh1)h=exlimh0(eh1)h

Use limh0(eh1)h=1, the derivative is y=ex.

Therefore, the derivative of y is, y=ex.

(c)

To determine

To find: The derivative of y=bx.

Expert Solution

Answer to Problem 2RCC

The derivative of y is, y=bxlnb.

Explanation of Solution

Calculation:

Obtain the derivative of y.

y=limh0f(x+h)f(x)h=limh0bx+hbxh=limh0bx(bh1)h=bxlimh0(bh1)h

Since limh0(bh1)h=lnb, the derivative is y=bxlnb.

Therefore, the derivative of y is y=bxlnb.

(d)

To determine

To find: The derivative of y=lnx.

Expert Solution

Answer to Problem 2RCC

The derivative of y=lnx is y=1x.

Explanation of Solution

Result used:

The value limh0(1+t)1t=et

Calculation:

Obtain the derivative of y.

y=limh0f(x+h)f(x)h=limh0ln(x+h)lnxh=limh0ln(x+hx)h        (ln(ab)=lnalnb)=limh01hln(1+hx)

Use lnam=mlna in the above,

y=limh0ln(1+hx)1h

Let t=hx then t tends to zero as h tends to zero.

y=limt0ln(1+t)1tx=limt0(ln(1+t)1t)1x=(limt0ln(1+t)1t)1x

Use the result stated above,

y=lne1x=1x         (Qlne=1)

Therefore, the derivative of y is y=1x.

(e)

To determine

To find: The derivative of y=logbx.

Expert Solution

Answer to Problem 2RCC

The derivative of y=logbx is, y=1xlnb.

Explanation of Solution

Calculation:

Obtain the derivative of y.

y=ddx(y)=ddx(logbx)=1xlnb    (Qddx(logax)=1xlna)

Therefore, the derivative of y is y=1xlnb.

(f)

To determine

To find: The derivative of y=sinx.

Expert Solution

Answer to Problem 2RCC

The derivative of y is, y=cosx.

Explanation of Solution

Calculation:

Obtain the derivative of y.

y=limh0f(x+h)f(x)h=limh0sin(x+h)sinxh=limh0sinxcosh+cosxsinhsinxh=limh0sinx(cosh1)+cosxsinhh

Simplify further,

y=limh0sinx(cosh1)h+limh0cosxsinhh=sinxlimh0(cosh1)h+cosxlimh0sinhh=sinx(0)+cosx(1)=cosx

Therefore, the derivative of y is y=cosx.

(g)

To determine

To find: The derivative of y=cosx.

Expert Solution

Answer to Problem 2RCC

The derivative of y is y=sinx.

Explanation of Solution

Calculation:

Obtain the derivative of y.

y=limh0f(x+h)f(x)h=limh0cos(x+h)cosxh=limh0cosxcoshsinxsinhcosxh=limh0cosx(cosh1)sinxsinhh

Simplify further,

y=limh0cosx(cosh1)hlimh0sinxsinhh=cosxlimh0(cosh1)hsinxlimh0sinhh=sinx(0)sinx(1)=sinx

Therefore, the derivative of y is y=sinx.

(h)

To determine

To find: The derivative of y=tanx.

Expert Solution

Answer to Problem 2RCC

The derivative of y is y=sec2x.

Explanation of Solution

Calculation:

Obtain the derivative of y.

y=limh0f(x+h)f(x)h=limh0tan(x+h)tanxh=limh0tanx+tanh1tanxtanhtanxh=limh0tanx+tanhtanx+tan2xtanh1tanxtanhh

Simplify further,

y=limh0(tanhh1+tan2x1tanxtanh)=limh0(sinhcoshhsec2x1tanxtanh)   (Q 1+tan2x=sec2x    tanh=sinhcosh)=limh0(sinhh1coshsec2x1tanxtanh)=sec2x

Therefore, the derivative of y is y=sec2x.

(i)

To determine

To find: The derivative of y=cscx.

Expert Solution

Answer to Problem 2RCC

The derivative of y is y=cscxcotx.

Explanation of Solution

Calculation:

Obtain the derivative of y.

y=limh0f(x+h)f(x)h=limh0csc(x+h)cscxh=limh01sin(x+h)1sinxh                   (Qcscx=1sinx)=limh01sinxcosh+cosxsinh1sinxh

Simplify further,

y=limh0sinxsinxcoshcosxsinh(sinxcosh+cosxsinh)sinxh=limh0sinx(1cosh)cosxsinhh1sin(x+h)sinx=[sinx(limh0(1cosh)h)(limh0cosxsinhh)](limh01sin(x+h)sinx)=[sinx(0)cosx(1)]1sin2x

That is, y=cosxsin2x.

y=cscxcotx

Therefore, the derivative of y is y=cscxcotx.

(j)

To determine

To find: The derivative of y=secx.

Expert Solution

Answer to Problem 2RCC

The derivative of y is y=secxtanx.

Explanation of Solution

Calculation:

Obtain the derivative of y.

y=limh0f(x+h)f(x)h=limh0sec(x+h)secxh=limh01cos(x+h)1cosxh                   (Qsecx=1cosx)=limh01cosxcoshsinxsinh1sinxh

Simplify further,

y=limh0cosxcosxcosh+sinxsinh(cosxcoshsinxsinh)cosxh=limh0cosx(1cosh)+sinxsinhh1cos(x+h)cosx=[cosx(limh0(1cosh)h)+(sinxlimh0xsinhh)](limh01cos(x+h)cosx)=[cosx(0)+sinx(1)]1cos2x

That is, y=sinxcos2x.

y=secxtanx

Therefore, the derivative of y is, y=secxtanx.

(k)

To determine

To find: The derivative of y=cotx.

Expert Solution

Answer to Problem 2RCC

The derivative of y is y=csc2x.

Explanation of Solution

Calculation:

Obtain the derivative of y.

y=limh0f(x+h)f(x)h=limh0cot(x+h)cotxh=limh0cos(x+h)sin(x+h)cosxsinxh                   (Qsecx=1cosx)=limh0sinxcos(x+h)cosxsin(x+h)hsin(x+h)sinx

Simplify further,

y=limh0sinx[cosxcoshsinxsinh]cosx[sinxcosh+cosxsinh]hsin(x+h)sinx=limh0sinxsin2hcos2xsinhhsin(x+h)sinx=limh0(sin2x+cos2x)sinhhsin(x+h)sinx=limh0sinhhsin(x+h)sinx               (Qsin2x+cos2x=1)

Apply the limit,

y=limh0(sinhh1sin(x+h)sinx)=(1)1sinxsinx=1sin2x=csc2x

Therefore, the derivative of y is y=csc2x.

(l)

To determine

To find: The derivative of y=sin1x.

Expert Solution

Answer to Problem 2RCC

The derivative of y is y=11x2.

Explanation of Solution

Calculation:

Consider the function y=sin1x.

That is siny=x and π2yπ2

Differentiate implicitly with respect x,

ddxsiny=ddxxcosydydx=1

dydx=1cosy (1)

Since cosy0 and π2yπ2.

cos2y=1sin2ycosy=1sin2ycosy=1x2        (Qsiny=x)

Substitute cosy=1x2 in equation (1),

dydx=11x2

Therefore, the derivative of y is y=11x2.

(m)

To determine

To find: The derivative of y=cos1x.

Expert Solution

Answer to Problem 2RCC

The derivative of y is y=11x2.

Explanation of Solution

Calculation:

Consider the function y=cos1x.

That is cosy=x and πy2π

Differentiate implicitly with respect x,

ddxcosy=ddxxsinydydx=1

dydx=1siny (1)

Since siny0 and πy2π.

sin2y=1cos2ysiny=1cos2ysiny=1x2        (Qcosy=x)

Substitute siny=1x2 in equation (1),

dydx=11x2

Therefore, the derivative of y is y=11x2.

(n)

To determine

To find: The derivative of y=tan1x.

Expert Solution

Answer to Problem 2RCC

The derivative of y is y=11+x2.

Explanation of Solution

Calculation:

Consider the function y=tan1x.

That is, tany=x

Differentiate implicitly with respect x,

ddxtany=ddxxsec2ydydx=1dydx=1sec2ydydx=11+tan2ydydx=11+x2   (Qtany=x)

Therefore, the derivative of y is y=11+x2.

(o)

To determine

To find: The derivative of y=sinhx.

Expert Solution

Answer to Problem 2RCC

The derivative of y is y=coshx.

Explanation of Solution

Calculation:

Obtain the derivative of y.

y=ddx(sinhx)=ddx(exex2) (Qsinhx=exex2)=12(ex+ex)                   =coshx

Therefore, the derivative of y is y=coshx.

(p)

To determine

To find: The derivative of y=coshx.

Expert Solution

Answer to Problem 2RCC

The derivative of y is y=sinhx.

Explanation of Solution

Calculation:

Obtain the derivative of y.

y=ddx(coshx)=ddx(ex+ex2) (Qcoshx=ex+ex2)=12(exex)                   =sinhx

Therefore, the derivative of y is y=sinhx.

(q)

To determine

To find: The derivative of y=tanhx.

Expert Solution

Answer to Problem 2RCC

The derivative of y is y=sech2x.

Explanation of Solution

Calculation:

Obtain the derivative of y.

y=ddx(tanhx)=ddx(sinhxcoshx)=coshxddx(sinhx)sinhxddx(coshx)cosh2x=coshx(coshx)sinhx(sinhx)cosh2x

Simplify further,

y=cosh2xsinh2xcosh2x=1cosh2x                 (Qcosh2xsinh2x=1)=sech2x                  (Q1coshx=sechx)

Therefore, the derivative of y is y=sech2x.

(r)

To determine

To find: The derivative of y=sinh1x.

Expert Solution

Answer to Problem 2RCC

The derivative of y is y=1x2+1.

Explanation of Solution

Calculation:

Consider the function y=sinh1x

That is, sinhy=x.

x=eyey2         (Qsinhx=exex2)x=eyey2(eyey)x=e2y12ey2xey=e2y1

Simplify further,

e2y2xey1=0(ey)22xey1=0

Solve the above quadric equation,

ey=2x+(2x)24(1)(1)2=2x+4x2+42=2x+2x2+12=x+x2+1

That is sinh1x=x+x2+1

Obtain the derivative y=sinh1x.

y=ddx(sinh1x)=ddx(ln(x+x2+1))=1x+x2+1ddx(x+x2+1)=1x+x2+1(1+12(x2+1)121(2x))

Simplify in terms and obtain the derivative,

y=1x+x2+1(1+x(x2+1)12)=1x+x2+1(1+x(x2+1)12)=1x+x2+1(x2+1+xx2+1)=1x2+1

Therefore, the derivative of y is y=1x2+1.

(s)

To determine

To find: The derivative of y=cosh1x.

Expert Solution

Answer to Problem 2RCC

The derivative of y is y=1x21.

Explanation of Solution

Calculation:

Consider the function y=cosh1x

That is, coshy=x.

x=eyey2         (Qcoshx=ex+ex2)x=ey+ey2(eyey)x=e2y+12ey2xey=e2y+1

Simplify further,

e2y2xey+1=0(ey)22xey+1=0

Solve the above quadric equation,

ey=2x+(2x)24(1)(1)2=2x+4x242=2x+2x212=x+x21

That is, cosh1x=x+x21.

Obtain the derivative y=cosh1x.

y=ddx(cosh1x)=ddx(ln(x+x21))=1x+x21ddx(x+x21)=1x+x21(1+12(x21)121(2x))

Simplify in terms and obtain the derivative,

y=1x+x21(1+x(x21)12)=1x+x21(1+x(x21)12)=1x+x21(x21+xx21)=1x21

Therefore, the derivative of y is y=1x21.

(t)

To determine

To find: The derivative of y=tanh1x.

Expert Solution

Answer to Problem 2RCC

The derivative of y is y=11x2.

Explanation of Solution

Calculation:

Consider the function y=tanh1x

That is, tanhy=x .

x=eyeyey+ey         (Qtanhx=ex+exex+ex)x=eyeyey+ey(eyey)x=e2y1e2y+1x(e2y+1)=e2y1

Simplify further,

x(e2y+1)e2y+1=0(x1)e2y+(x+1)=0e2y=(x+1)(x1)

Take natural logarithm on both sides,

lne2y=ln(x+1)(x1)2y=ln(x+1)ln[(x1)]y=12[ln(x+1)ln(1x)]

That is, tanh1x=12[ln(x+1)ln(1x)] .

Obtain the derivative y=cosh1x.

y=ddx(tanh1x)=ddx(12[ln(x+1)ln(1x)])=12[1x+111x(1)]=12[1x+x+11x2]=11x2

Therefore, the derivative of y is y=11x2.

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