Chapter 5 Higher Order Derivatives-I
Also called as successive derivatives.
Formulas for some typical functions
| Function | nth derivative | Remarks |
|---|---|---|
| y=xm | yn=m!(m−n)!xm−n | m is a positive integer |
| y=xm | yn=n! | if m=n be a positive integer |
| y=(ax+b)m | yn=m!(m−n)!an(ax+b)m−n | m is a positive integer, n<m |
| y=logx | yn=(−1)n−1(n−1)!xn | |
| y=log(ax+b) | yn=(−1)n−1(n−1)!an(ax+b)n | |
| y=1x | yn=(−1)nn!xn+1 | |
| y=1ax+b | yn=(−1)nn!an(ax+b)n+1 | |
| y=amx | yn=mnamx(loga)n | |
| y=eax | yn=aneax | |
| y=ax | yn=ax(logea)n | |
| y=sin(ax+b) | yn=ansin(ax+b+nπ2) | |
| y=cos(ax+b) | yn=ancos(ax+b+nπ2) |
| Function | nth derivative | Remarks |
|---|---|---|
| y=eaxsinbx | yn=(a2+b2)n/2eaxsin(bx+ntan−1ba) | |
| y=eaxcosbx | yn=(a2+b2)n/2eaxcos(bx+ntan−1ba) | |
| y=1x2−a2 | yn=(−1)nn!2a[1(x−a)n+1−1(x+a)n+1] | |
| y=1x2+a2 | yn=(−1)nn!an+2sinn+1θsin(n+1)θ | where θ=tan−1(ax) |
| y=tan−1x | yn=(−1)n−1(n−1)!sinnθsinnθ | where θ=tan−1(1x) |
5.1 Exercise 3 (i)
5.1.1 Question 1
Find yn of the following:
- y=(a−bx)m
y1=−m(a−bx)m−1by2=m(m−1)(a−bx)m−2b2y3=−m(m−1)(m−2)(a−bx)m−3b3⋮yn=(−1)nm(m−1)(m−2)…(m−n+1)bn(a−bx)m−n
- y=1a−x
y1=−1(a−x)−2×−1=(a−x)−2y2=−2(a−x)−3×−1=2(a−x)−3y3=−6(a−x)−4×−1=6(a−x)−4⋮yn=n!(a−x)n+1
- y=x2n
y1=2nx2n−1y2=2n(2n−1)x2n−2y3=2n(2n−1)(2n−2)x2n−3⋮yn=2n(2n−1)(2n−2)(2n−3)(2n−4)…(2n−n+1)x2n−n=2n(2n−1)(2n−2)(2n−3)(2n−4)…(n+1)xn=2n(2n−1)(2n−2)(2n−3)(2n−4)…(n+1)n(n−1)(n−2)(n−3)…3.2.1n(n−1)(n−2)(n−3)…3.2.1xn
Separating even and odd factors, we write
yn=[2n(2n−2)(2n−4)(2n−8)…6.4.2][(2n−1)(2n−3)…5.3.1]n!xnyn=2n[n(n−1)(n−2)…3.2.1][1.3.5…(2n−3)(2n−1)]n!xnyn=2n[1.3.5…(2n−3)(2n−1)]xn
- y=√x
y1=12x−1/2y2=−12×2x−3/2y3=1×32×2×2x−5/2y4=−1×3×52×2×2×2x−7/2y5=1×3×5×72×2×2×2×2x−9/2⋮yn=(−1)n−11.3.5.7.9…(2n−3)2nxn−1/2
- y=1√x
y1=−12x−3/2y2=1×32×2x−5/2y3=−1×3×52×2×2x−7/2y4=1×3×5×72×2×2×2x−9/2⋮yn=(−1)n1.3.5.7…(2n−1)2nxn+1/2
- y=103−2x
Taking log both sides,
logy=(3−2x)log10=3log10−2xlog10
Differentiating w.r.t x,
1ydydx=0−2log10y1=−2log10y=−2log10×103−2xy2=−2log10y1=2log10×2log10×103−2xy3=−2log10y2=−2log10×2log10×2log10×103−2xy4=−2log10y3=2log10×2log10×2log10×2log10×103−2x⋮yn=(−1)n2n(log10)n103−2x
- y=xa+bx
Breaking into partial fractions,
y=1b−ab(a+bx)
Differentiating w.r.t x,
y1=0−ab×(−1)(a+bx)−2×b=(−1)2a(a+bx)2y2=(−1)2a×(−2)(a+bx)−3×b=(−1)32ab(a+bx)3y3=(−1)32ab×(−3)(a+bx)−4×b=(−1)46ab2(a+bx)4y4=(−1)46ab2×(−4)(a+bx)−5×b=(−1)524ab3(a+bx)5=⋮yn=(−1)n+1n!abn−1(a+bx)n+1
- y=xnx−1
Let x−1=t, then x=t+1.
The equation can be written as
y=(t+1)nt
Using binomial expansion,
y=nC0tn.1+nC1tn−1.12+nC2tn−2.13+…+nCn.1nty=tn+nC1tn−1.1+nC2tn−2.1+…+1ty=tn−1+nC1tn−2+nC2tn−3+…+1ty=(x−1)n−1+nC1(x−1)n−2+nC2(x−1)n−3+…+1(x−1)
The nth derivative is
yn=dnydxn[(x−1)n−1+nC1(x−1)n−2+nC2(x−1)n−3+…+1(x−1)]yn=0+0+0+…+(−1)nn!(x−1)n+1=(−1)nn!(x−1)n+1
- y=exsinxsin2x
sinxsin2x can be written as
=12[cos(x−2x)−cos3x]=12[cos(−x)−cos3x]=12[cosx−cos3x]
Original equation can be written as thus
y=exsinxsin2x=ex×12[cosx−cos3x]=12[excosx−excos3x]
The nth derivative is
=12[(12+12)n/2excos(x+ntan−1(1/1))−(12+32)n/2excos(3x+ntan−1(3/1))]=12[2n/2excos(x+nπ4)−10n/2excos(3x+ntan−1(3/1))]=12[2n/2excos(x+nπ4)−10n/2excos(3x+ntan−13)]=12ex[2n/2cos(x+nπ4)−10n/2cos(3x+ntan−13)]
- y=e3xsin4x
This equation is of the form y=eaxsinbx where a=3 and b=4. So from the table above, we have,
yn=(a2+b2)n/2eaxsin(bx+ntan−1ba)yn=(32+42)n/2e3xsin(4x+ntan−143)=(25)n/2e3xsin(4x+ntan−143)yn=5ne3xsin(4x+ntan−143)
- y=exsin2x
The equation can be written as y=exsinxsinx.
We know from trigonometry,
sinαsinβ=12[cos(α−β)−cos(α+β)]
So
sinxsinx=12[cos(x−x)−cos(x+x)]=12[1−cos2x]
Thus the given equation can be written as
y=exsinxsinxy=ex(12[1−cos2x])y=12[ex−excos2x]
And its nth derivative is
yn=12[ex−(12+22)n/2e1xcos(2x+ntan−121)]yn=12[ex−5n/2excos(2x+ntan−12)]
5.1.2 Question 2
Find the nth derivatives of the following functions:
- y=1x2+16
This equation can be written as y=1x2+42 which is of the form y=1x2+a2, so
yn=(−1)nn!4n+2sinn+1θsin(n+1)θ
where θ=tan−1(4x).
- y=xx2+a2
Breaking the equation into partial fractions for simplicity of calculating nth derivative,
y=xx2−(ia)2,where i=√−1=x(x+ia)(x−ia)=12[1x+ia+1x−ia]yn=12[Dn(1x+ia)+Dn(1x−ia)]
The nth derivative of 1ax+b is (−1)nn!an(ax+b)n+1.
yn=12[(−1)nn!(x+ia)n+1+(−1)nn!(x−ia)n+1]=(−1)nn!2[(x+ia)−(n+1)+(x−ia)−(n+1)]
Lets put x=rcosθ and a=rsinθ.
=(−1)nn!2[(rcosθ+irsinθ)−(n+1)+(rcosθ−irsinθ)−(n+1)]
Applying De-Moivre’s theorem,
yn=(−1)nn!2[r−(n+1){cos(−(n+1)θ)+isin(−(n+1)θ)}+r−(n+1){cos(−(n+1)θ)−isin(−(n+1)θ)}]yn=(−1)nn!2rn+1[2cos(n+1)θ]yn=(−1)nn!rn+1[cos(n+1)θ]
cos(−θ)=cosθ
Putting r=asinθ,
yn=(−1)nn!an+1sinn+1θcos(n+1)θ
where θ=tan−1ax.
- y=1x2+x+1
The x values of denominator from the formula −b±√b2−4ac2a are −1+√3i2 and −1−√3i2.
So the equation can be resolved into partial fractions as follows,
y=1[x−(−1+√3i2)][x−(−1−√3i2)]=4[(2x+1)−√3i][(2x+1)+√3i]y=42√3i[1(2x+1)−√3i−1(2x+1)+√3i]y=2√3i[1(2x+1)−√3i−1(2x+1)+√3i]
yn=2√3i[Dn(1(2x+1)−√3i)−Dn(1(2x+1)+√3i)]
yn=2√3i[(−1)nn!2n(2x+1−√3i)n+1−(−1)nn!2n(2x+1+√3i)n+1]
The nth derivative of 12x+a+b is (−1)n2nn!(2x+a+b)n+1.
Lets put 2x+1=rcosθ and √3=rsinθ, so that tanθ=√32x+1.
yn=(−1)n2n+1n!√3i[(rcosθ−irsinθ)−(n+1)−(rcosθ+irsinθ)−(n+1)]
Applying De-Moivre’s theorem,
yn=(−1)n2n+1n!√3i×r−(n+1)[(cosθ−isinθ)−(n+1)−(cosθ+isinθ)−(n+1)]=(−1)n2n+1n!√3i×r−(n+1)[{cos(−(n+1)θ)−isin(−(n+1)θ)}−{cos(−(n+1)θ)+isin(−(n+1)θ)}]=(−1)n2n+1n!√3i×r−(n+1)[cos(n+1)θ+isin(n+1)θ−cos(n+1)θ+isin(n+1)θ]
yn==(−1)n2n+1n!r−(n+1)√3i[2isin(n+1)θ]yn==(−1)n2n+2n!√3r(n+1)[sin(n+1)θ]
Substituting r=√3sinθ, we get
yn=(−1)n2n+2n!√3(√3sinθ)n+1[sin(n+1)θ]yn=(−1)n2n+2n!(√3)n+2sinn+1θsin(n+1)θ
where θ=tan−1(√32x+1).
- y=tan−11+x1−x
Lets assume x=tanθ.
Then equation becomes,
y=tan−1tanπ/4+tanθ1−tanπ/4tanθy=tan−1tan(π/4+θ)y=π4+θ
So nth derivative is,
yn=Dn(π4)+Dn(tan−1x)yn=0+Dn(tan−1x)yn=(−1)n−1(n−1)!sinnϕsinnϕ
where ϕ=tan−11x.
The nth derivative of tan−1x is (−1)n−1(n−1)!sinnθsinnθ where θ=tan−11x.
- y=1(x2+a2)(x2+b2)
Breaking the equation into derivable known partial fractions,
y=1b2−a2[1x2+a2−1x2+b2]yn=1b2−a2[Dn(1x2+a2)−Dn(1x2+b2)]=1b2−a2[(−1)nn!an+2sinn+1θsin(n+1)θ−(−1)nn!bn+2sinn+1ϕsin(n+1)ϕ]=(−1)nn!b2−a2[sinn+1θsin(n+1)θan+2−sinn+1ϕsin(n+1)ϕbn+2]yn=(−1)nn!a2−b2[sinn+1ϕsin(n+1)ϕbn+2−sinn+1θsin(n+1)θan+2]
where θ=tan−1(ax) and ϕ=tan−1(bx)
- y=cot−1xa
Differentiating w.r.t x,
y1=−a2a2+x2×1ay1=(−1)×ax2+a2
Differentiating the above equation n−1 times gives nth derivative. We can use general formula for yn for expression 1x2+a2.
So in our case,
y1+n−1=(−1)Dn−1(ax2+a2)y1+n−1=(−1)a[(−1)n−1(n−1)!an−1+2sinn−1+1θsin(n−1+1)θ]yn=(−1)a[(−1)n−1(n−1)!an+1sinnθsin(n)θ]=(−1)n(n−1)!ansinnθsin(n)θ
where θ=tan−1(ax)
- y=tan−1√1+x2−1x
Lets assume x=tanθ.
So the equation becomes,
y=tan−1√1+tan2θ−1tanθ=tan−1secθ−1tanθ=tan−11−cosθsinθ=tan−12sin2θ/22sinθ/2cosθ/2=tan−1tanθ2y=θ2y=12θy=12tan−1x
So nth derivative is,
yn=12(−1)n−1(n−1)!sinnθsinnθ
where θ=tan−11x.
5.1.3 Question 3
If y=1xx, show that y2(1)=0.
Taking log both sides,
logy=−xlogx
Differentiating w.r.t x
1ydydx=−x1x−logx=−1−logxy1=−y(1+logx)
Again differentiating w.r.t x,
dy1dx=−[y1(1+logx)+yx]y2=−[−y(1+logx)(1+logx)+yx]y2=y[(1+logx)2−1x]y2=1xx[(1+logx)2−1x]
So,
y2(1)=11[(1+0)2−1]=1(1−1)=0
Hence proved.
5.1.4 Question 4
If ax2+2hxy+by2=1, show that d2ydx2=h2−ab(hx+by)3.
dydx=−fxfy=−2ax+2hy2hx+2byy1=−ax+hyhx+byy2=−[(hx+by)(a+hy1)−(ax+hy)(h+by1)](hx+by)2=−[(hx+by)(a−h(ax+hy)hx+by)−(ax+hy)(h−b(ax+hy)hx+by)](hx+by)2=−[(hx+by)(ahx+aby−ahx−h2yhx+by)−(ax+hy)(h2x+bhy−abx−bhyhx+by)](hx+by)2=−[(hx+by)(aby−h2y)−(ax+hy)(h2x−abx)](hx+by)2×(hx+by)=x(ax+hy)(h2−ab)−y(ab−h2)(hx+by)(hx+by)3=x(ax+hy)(h2−ab)+y(h2−ab)(hx+by)(hx+by)3=(h2−ab)(ax2+hxy+hxy+by2)(hx+by)3=(h2−ab)(ax2+2hxy+by2)(hx+by)3d2ydx2=h2−ab(hx+by)3, as ax2+2hxy+by2=1
5.1.5 Question 5
If y=e−pxcosqx, prove that y2+2py1+(p2+q2)y=0.
Differentiating w.r.t x,
y1=−qe−pxsinqx−pe−pxcosqx=−qe−pxsinqx−pyy2=−qddx(e−pxsinqx)−py1y2=−q[qe−pxcosqx−pe−pxsinqx]−py1=−q[qy+py1+pyq]−py1y2=−qq[q2y+py1+p2y]−py1y2+(p2+q2)y+2py1=0
5.1.6 Question 6
If y=x2n, where n is a positive integer, show that yn=2n[1.3.5…(2n−3)(2n−1)]xn.
y=x2ny1=2nx2n−1y2=2n(2n−1)x2n−2y3=2n(2n−1)(2n−2)x2n−3⋮yn=2n(2n−1)(2n−2)(2n−3)(2n−4)…(2n−n+1)x2n−n=2n(2n−1)(2n−2)(2n−3)(2n−4)…(n+1)xn=2n(2n−1)(2n−2)(2n−3)(2n−4)…(n+1)n(n−1)(n−2)(n−3)…3.2.1n(n−1)(n−2)(n−3)…3.2.1xn
Separating even and odd factors, we write
yn=[2n(2n−2)(2n−4)(2n−8)…6.4.2][(2n−1)(2n−3)…5.3.1]n!xnyn=2n[n(n−1)(n−2)…3.2.1][1.3.5…(2n−3)(2n−1)]n!xnyn=2n[1.3.5…(2n−3)(2n−1)]xn
5.1.7 Question 7
If y=sinmx+cosmx, prove that
- y2+m2y=0
y1=mcosmx−msinmxy2=−m2sinmx−m2cosmx=−m2(sinmx+cosmx)=−m2yy2+m2y=0
- yn=mn{1+(−1)nsin2mx}1/2
We have, y=sinmx+cosmx
dnydxn=Dn(sinmx)+Dn(cosmx)yn=mnsin(mx+0+nπ2)+mncos(mx+0+nπ2)=mn[sin(mx+nπ2)+cos(mx+nπ2)]
Squaring and taking square root,
yn=mn[√{sin(mx+nπ2)+cos(mx+nπ2)}2]=mn[√1+2sin(mx+nπ2)cos(mx+nπ2)]=mn[√1+sin2(mx+nπ2)]=mn[√1+sin(2mx+nπ)]=mn[√1+sin2mxcosnπ+cos2mxsinnπ]=mn{1+(−1)nsin2mx}1/2
because cosnπ=(−1)n and sinnπ=0.
Thus,
yn=mn{1+(−1)nsin2mx}1/2