Class 11 Maths Ncert Solutions Ex 13.2

Class 11 Maths Ncert Solutions Chapter 13 Ex 13.2

Q1. Find derivative for x2 – 2 at x = 10

 

Soln:

Let f(x) = x2 – 2

Accordingly,

f(10)=limh0f(10+h)f(10)h

 

=limh0[(10+h)22](1022)h

 

=limh0102+2.10.h+h22102+2h

 

=limh020h+h2h

 

=limh0(20+h)=(20+0)=0

 

Hence, derivative for x2 – 2 at x = 10 is 20

 

 

Q2. Find derivative

99x at x = 100

 

Soln:

Let f(x) = 99x,

Accordingly,

f(100)=limh0f(100+h)f(100)h

 

=limh099(100+h)99(100)h

 

=limh099×100+99h99×100h

 

=limh099hh

 

=limh0(99h)=99

 

Hence, derivative for 99x at x = 100 is 99

 

 

Q3. Find derivative

x at x = 1

 

Soln:

Let f(x) = x

Accordingly,

f(1)=limh0f(1+h)f(1)h

 

=limh0(1+h)1h

 

=limh0hh

 

=limh0(1)

 

= 1

Hence, derivative for x at x = 1 is 1

 

 

Q4. Using first principle find derivative

(i) x3 – 27

 

(ii) (x – 1) (x – 2)

 

(iii) 1x2

 

(iv) x+1x1

Soln:

(i) Let f(x) = x3 – 27

From first principle,

f(x)=limh0f(x+h)f(x)h

 

=limh0[(x+h)327](x327)h

 

=limh0x3+h3+3x2h+3xh2x3h

 

=limh0h3+3x2h+3xh2h

 

=limh0(h2+3x2+3xh)

 

= 0 + 3x2 + 0 = 3x2

 

(ii) Let f(x) = (x – 1)(x – 2)

From first principle,

f(x)=limh0f(x+h)f(x)h

 

=limh0(x+h1)(x+h2)(x1)(x2)h

 

=limh0(x2+hx2x+hx+h22hxh+2)(x22xx+2)h

 

=limh0(hx+hx+h22hh)h

 

=limh0(2hx+h23h)h

 

=limh0(2x+h3)

 

(2x+h3)

 

= 2x – 3

 

(iii) Let f(x)=1x2

From first principle

f(x)=limh0f(x+h)f(x)h

 

=limh01(x+h)21x2h

 

=limh01h[x2(x+h)2x2(x+h)2]

 

=limh01h[x2x2h22hxx2(x+h)2]

 

=limh01h[h22hxx2(x+h)2]

 

=limh0[h2hxx2(x+h)2]

 

02xx2(x+0)2=2x3

 

(iv) Let f(x)=x+1x1

From first principle,

=limh0(x+h+1x+h1x+1x1)h

 

=limh01h[(x1)(x+h+1)(x+1)(x+h1)(x1)(x+h1)]

 

=limh01h[(x2+hx+xxh1)(x2+hxx+x+h1)(x1)(x+h+1)]

 

=limh01h[2h(x1)(x+h1)]

 

=limh0[2(x1)(x+h1)]

 

=2(x1)(x1)=2(x1)2

 

 

Q5. Prove f(1)=100f(0)

For the function f(x)=x100100+x9999++x22+x+1

 

Soln:

Given function

f(x)=x100100+x9999++x22+x+1

 

ddxf(x)=ddx[x100100+x9999++x22+x+1]

 

ddxf(x)=ddx(x100100)+ddx(x9999)++ddx(x22)+ddx(x)+ddx(1)

 

Using theorem ddx(xn)=nxn1, we get

 

ddxf(x)=100x99100+99x9899+....+2x2+1+0

 

=x99+x98+.+x+1

 

∴ ddxf(x)=x99+x98+.+x+1

 

At x = 0

 

f'(0) = 1

 

At x = 1,

 

f'(1) = 199 + 198 + . . . . + 1 + 1 = [1 + 1 +…… + 1 +1 ]100 terms = 1 x 100 = 100

 

Hence,  f’(1) = 100 x f1 (0)

 

 

Q6. For a real number ‘a’ find the derivative of xn + axn – 1 + a2xn – 2 + . . . .+ an – 1 x + an

Soln:

Let f(x)=xn+axn1+a2xn2+...+an1x+an

 

f(x)=ddx(xn+axn1+a2xn2+...+an1x+an)

 

=ddx(xn)+addx(xn1)+a2ddx(xn2)+...+an1ddx(x)+anddx(1)

 

Using ddxxn=nxn1, we have

 

f(x)=nxn1+a(n1)xn2+a2(n2)xn3++an1+an(0)

 

=nxn1+a(n1)xn2+a2(n2)xn3++an1

 

 

Q7. Find the derivative for

(i) (x – m)(x – n)

 

(ii) (ax2 + b)2

 

(iii) xaxb

Soln:

(i) Let f(x) = (x – m)(x – n)

 

f(x)=x2(m+n)x+mn

 

f(x)=ddx(x2(m+n)x+mn)

 

=ddx(x2)(m+n)ddx(x)+ddx(mn)

 

Using =ddx(xn)=nxn1, we get

 

f(x)=2x(m+n)+0=2xmn

 

(ii) Let f(x) = (ax2 + b)2

 

f(x)=a2x2+2abx+b2

 

f(x)=ddx(a2x4+2abx2+b2)=a2ddx(x4)+2abddx(x2)+ddx(b2)

 

Using ddxxn=nxn1, we have

 

f(x)=a2(4x3)+2ab(2x)+b2(0)

 

= 4a2x3 + 4abx

 

= 4ax(ax2 + b)

 

(iii) Let f(x)=(xa)(xb)

 

f(x)=ddx(xaxb)

 

Quotient rule,

f(x)=(xb)ddx(xa)(xa)ddx(xb)(xb)2

 

=(xb)(1)(xa)(1)(xb)2

 

=xbx+a(xb)2

 

=ab(xb)2

 

 

Q8. If a is constant find derivative of xnanxa

Soln:

Let xnanxa

 

f(x)=ddx(xnanxa)

 

By Question rule

f(x)=(xa)ddx(xnan)(xnan)ddx(xa)(xa)2

 

=(xa)(nxn10)(xnan)(xa)2

 

=nxnanxn1xn+an(xa)2

 

 

Q9.

(i) 2x34

 

(ii) (5x3 + 3x – 1) (x – 1)

 

(iii) x-3 (5 + 3x)

 

(iv) x5 (3 – 6x-9)

 

(v) x-4 (3 – 4x-5)

 

(vi) =2x+1x23x1

 

Soln.

(i) Let f(x)=2x34

 

f(x)=ddx(2x34)

 

=2ddx(x)ddx(34)

 

= 2 – 0

 

= 2

 

(ii) Let f(x) = (5x3 + 3x – 1) (x – 1)

By Leibnitz product rule,

f(x)=(5x3+3x1)ddx(x1)+(x1)ddx(5x3+3x1)

 

= (5x3 + 3x – 1) (1) + (x – 1)(5.3x2 + 3 – 0)

 

= (5x3 + 3x – 1) + (x – 1)(15x2 + 3)

 

= 5x3 + 3x – 1 + 15x3 + 3x – 15x2 – 3

 

= 20x3 – 15x2 + 6x – 4

 

(iii) Let f(x) = x-3 (5 + 3x)

 

f(x)=x3ddx(5+3x)+(5+3x)ddx(x3)

 

=x3(0+3)+(5+3x)(3x31)

 

=x3(3)+(5+3x)(3x4)

 

= 3x-3 – 15x-4 – 9x-3

 

= -6x-3 – 15x-4

 

=3x3(2+5x)

 

=3x3x(2x+5)

 

=3x4(2x+5)

 

(iv) let f(x) = x5 (3 – 6x-9)

 

From Leibnitz product rule,

f(x)=x5ddx(36x9)+(36x9)ddx(x5)

 

=x5{06(9)x91}+(36x9)(5x4)

 

=x5(54x10)+15x430x5

 

= 54x-5 + 15x4 – 30x-5

 

= 24x-5 + 15x4

 

= =15x4+24x5

 

(v) Let f(x) = x-4 (3 – 4x-5)

 

From Leibnitz product rule,

f(x)=x4ddx(34x5)+(34x5)ddx(x4)

 

=x4{04(5)x51+(34x5)(4)x41}

 

= x-4 (20x-6) + (3 – 4x-5)(-4x-5)

 

= 36x-10 – 12x-5

 

=12x5+36x10

 

(iv) Let f(x)=2x+1x23x1

f(x)=ddx(2x+1)ddx(x23x1)

 

From quotient rule,

f(x)=[(x+1)ddx(2)2ddx(x+1)(x+1)2][(3x1)ddx(x2)x2ddx(3x1)(3x1)2]

 

=[(x+1)(0)2(1)(x+1)2][(3x1)(2x)(x2)(3)(3x1)2]

 

=2(x+1)2[6x22x3x2(3x1)2]

 

=2(x+1)2[3x22x(3x1)2]

 

=2(x+1)2x(3x2)(3x1)2

 

 

Q10. Using first principle find derivative of cosx

Soln:

Let f(x)=cosx.

 

According to first principle

f(x)=limh0f(x+h)f(x)h

 

=limh0cos(x+h)cosxh

 

=limh0[cosxcoshsinxsinhcosxh]

 

=limh0[cosx(1cosh)sinxsinhh]

 

=limh0[cosx(1cosh)hsinxsinhh]

 

=cosx(limh01coshh)sinxlimh0(sinhh)

 

= =cosx(0)sinx(1)   [limh01coshh=0andlimh0sinhh=1]

 

= – sin x

 

f’(x) = – sin x

 

 

Q11.

(i) sin x cos x

 

(ii) sec x

 

(iii) 5 sec x + 4 cos x

 

(iv) cosec x

 

(v) 3cot x + 5cosec x

 

(vi) 5sin x – 6cos x + 7

 

(vii) 2tan x – 7sec x

Soln:

(i) Let f’(x) = sin x cos x

 

According to first principle,

f(x)=limh0f(x+h)f(x)h

 

=limh0sin(x+h)cos(x+h)sinxcosxh

 

=limh012h[2sin(x+h)cos(x+h)2sinxcosx]

 

=limh012h[sin2(x+h)sin2x]

 

=limh012h[2cos2x+2h+2x2.sin2x+2h2x2]

 

=limh01h[cos4x+2h2sin2h2]

 

limh01h[cos(2x+h)sinh]

 

limh0cos(2x+h)limh0sinhh

 

= cos(2x + 0). 1

 

= cos 2x

 

(ii) Let f(x) = sec x

According to first principle,

f(x)=limh0f(x+h)f(x)h

 

=limh0sec(x+h)secxh

 

=limh01h[1cos(x+h)1cosx]

 

=limh01h[cosxcos(x+h)cosxcos(x+h)]

 

=1cosxlimh01h[2sin(x+x+h2)sin(xxh2)cos(x+h)]

 

=1cosxlimh01h[2sin(2x+h2)sin(h2)cos(x+h)]

 

=1cosxlimh0sin(2x+h2)sin(h2)(h2)cos(x+h)

 

=1cosxlimh0sin(h2)(h2).limh0sin(2x+h2)cos(x+h)

 

=1cosx.1.sinxcosx

 

=secxtanx

 

(iii)

Let f(x) = 5 sec x + 4 cos x

 

According to first principle,

f(x)=limh0f(x+h)f(x)h

 

limh05sec(x+h)+4cos(x+h)[5 secx+4 cosx]h

 

=5limh0[sec(x+h)secx]h+4limh0[cos(x+h)cosx]h

 

=5limh01h[1cos(x+h)1cosx]+4limh01h[cos(x+h)cosx]

 

=5limh01h[cosxcos(x+h)cosxcos(x+h)]+4limh01h[cosxcoshsinxsinhcosx]

 

=5cosxlimh01h[2sin(x+x+h2)sin(xxh2)cos(x+h)]+4limh01h[cosx(1cosh)sinxsinh]

 

=5cosxlimh01h[2sin(2x+h2)sin(h2)cos(x+h)]+4[cosxlimh0(1cosh)hsinxlimh0sinhh]

 

=5cosx.limh0sin(2x+h2).sin(h2)h2cos(x+h)+4[(cosx).(0)(sinx).1]

 

=5cosx.[limh0sin(2x+h2)cos(x+h).limh0sin(h