Limits and Derivatives

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New answer posted•a year ago
S
a year ago

x + 1 − 2 l o g 2 ( 3 + 2 x ) + 2 l o g 4 ( 1 0 − 2 − x ) = 0

x + 1 + l o g 2 [ 1 0 . 2 x − 1 ( 3 + 2 x ) 2 ] − x = 0

( 2 x ) 2 − 1 4 . 2 x + 1 1 = 0

2x + y

x 1 + x 2 = l o g 2 ( 4 9 − 1 5 2 4 )

x 1 + x 2 = l o g 2 1 1

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New answer posted•a year ago
S
a year ago

(i) f(x)=sin x cos x

So, f?(x)=limh?0f(x+h)?f(x)h

=limh?0sin(x+h)cos(x+h)?sinxcosxh

=limh?012h×[2sin(x+h)cos(x+h)?2sinxcosx]

=limh?012h[sin2(x+h)?sin2x]

=limh?012h[2cos2(x+h)+2x2sin2(x+h)?2x2]

=limh?01h[cos(2x+h)sinh]

=limh?0cos(2x+h)×limh?0sinhh

=cos(2x+0)

=cos2x

(ii) f(x)=secx

So, f?(x)=limh?0f(x+h)?f(x)h

=limh?01h[sec(x+h)?secx]

=limh?01h[1cos(x+h)?1cosx]

=limh?01h[cosx?cos(x+h)cos(x+h)cosx]

=limh?01h[?2sin(x+x+h2)sin(x?(x+h)2)cos(x+h)cosx]

=limh?01h[?2sin(2x+h2)sin(?h/2)cos(x+h)cosx]

=limh?0(?1sin(2x+h2)cos(x+h)cosx×limh?0(?1)sinh/2h/2

=sinxcosx?cosx×1

=tanx?secx.

(iii) Given f(x)=5 sec x+4 cosx.

So, f?(x)=limh?0f(x+h)?f(x)h

=limh?05sec(x+h)+4cos(x+h)?[5secx+4cosx]h.

=limh?05h[sec(x+h)?secx]+limh?04h[cos(x+h)?cosx]

=limh?05h[1cos(x+h)?1cosx]+limh?04h[?2sin(x+h+x2)sin(x+h?x2)]

=limh?05h[cosx?cos(x+h)cos(x+h)(cosx)]+limh?04h[?2sin(2x+h2)sinh2]

=limh?05h[?2sin(2x+h2)sin(?h/2)cos(x+h)cosx]?4limh?0sin(2x+h2)limh?0sinh/2h/2

=sin(2x+02)cos(x+0)cosx×1?4sin(2x2)

=5sinxcosx?1cosx?4sinx

=5tanx?secx?4sinx

(iv) Given f(x)=cosecx

f?(x)=limh?0f(x+h)?f(x)h

=limh?01h[cosec(x+h)?cosecx]

=limh?01h[1sin(x+h)?1sinx]

=limh?01h[sinx?sin(x+h)sin(x+h)sinx]

=limh?01h[2cos(x+x+h2)sin(x?(x+h)2)]sin(x+h)sinx]

=limh?01h[2cos(x+x+h2)sin(x?(x+h)2)sin(x+h)sinx]

=limh?01h[2cos(2x+h2)sin(?h/2)]sin(x+h)sinx]

=limh?0cos(2x+h2)sin(x+h)sinx×(?1)sin(2)h/2)

=cos(2x+02)sin(x+0)sinx×(?1)

=?cosxsinx×1sinx

=?cotx?cosecx

(v) Given,f(x)=3 cot x+5cosecx.

So, f?(x)=limh?0f(x+h)?f(x)h =2cos(x+0)cosx×1+7sin(2x+02)×(?1)

=limh?03cot(x+h)+5cosec(x+h)?[3cotx+5cosx)

h?03h[cot(x+h)?cotx]+limh?0

=?5h[cosec(x+h)?cosecx]

=limh?03h[cos(x+h)sin(x+h)?cosxsinx]+limh?05h[1sin(x+h)?1sinx]

=limh?03h[cos(x+h)sinx?cosxsin(x+h)sin(x+h)sinx]+limh?05h[sinx?sin(x+h)sin(x+h)sinx]

=limh?03h[sin(x?(x+h))sin(x+h)sinx+limh?05h[2cos(x+x+h2)sin(x?(x+h)2)sin(x+h)sinx

=limh?03sin(x+h)sinx×limh?0(?1)sinhh+limh?05h[2cos(2x+h2)sin(?h/2)sin(x+h)sinx

=3sinx?sinx×(?1)+5limh

...Read more

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New answer posted•a year ago
S
a year ago

29.

Given, f (x) = (x-a1) (x-a2)… (x-an)

So,  limx→a1f (x)=limx→a1 (x−a1)limx→a1 (x−a2)? limx→a1 (x−an)

= (a1-a1) (a1-a2) … (a1-an)

= 0 (a1-a2) … (a1-an)

 = 0

And limx→af (x)=limx→a (x−a1)limx→a (x−a2)…limx→a (x−an)

= (a-a1) (a-a2) … (a-an)

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New answer posted•a year ago
S
a year ago

17. Kindly go through the solution

==limx→02sin2x2sin2x2

=limx→0 (sinxx)2×x2limx→0 (sinx2x2)2x22

= (1)2×x2×4 (1)2×x2

= 4

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New answer posted•a year ago
S
a year ago

18.  limx? 0ax+xcosxbsinx=limx? 0x (a+cosx)bsinx

=1b* (a+cos0)1

=a+1b

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New answer posted•a year ago
S
a year ago

73. Given, f (x) = xsinnx

So, f?(x) = (sinxxdxdx)−(xddxsinxx)(sinxx)2

= sinxx−x·n(sinx)x−1ddxsinx(sinx)2x

=(sinx)n−x·x(sinx)n−1·cosx(sinx)2n

=(sinx)n−1[sinx−nx·cosx](sinx)2n

=sinx−n·x·cosx(sinx)2n−(n−1)

=sinx−nxcosx(sinx)n+1

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New answer posted•a year ago
S
a year ago

72. Given, f (x) = (x + sec x) (x tan x)

So, f?(x) = (x + see x) ddx(x−tanx)+(x−tanx)ddx(x+secx).

=(x+secx)(dxdx−ddxtanx)+(x−tanx)(dxdx+ddxsecx)

Let g(x) = tan x.

So, g?(x) =

 

=limh→01h[tan(x+h)−tanx]

=limh→01h[sin(x+h)cos(x+h)−sinxcosx]

=limh→01h[sin(x+h)cosx−sinxcos(x+h)cos(x+h)cosx]

=limh→01h[sin(x+h−x)cos(x+h)cosx]

=limh→01cos(x+h)cosx×limh→0sinhh

 

=1cos2x=sec2x . _____ (2)

And P(x) = see x.

=limh→0sec(x+h)−secxh

=limh→01h[1cos(x+h)−1cosx]

=limh→01h[cosx−cos(x+h)cos(x+h)cosx]

=limh→01h[−2sin(x+x+h2)sin(x−(x+h)2)cos(x+h)cosx]

 

=sinxcos2x = tan x sec x. ____ (3)

   Putting (2) and (3) in (1) we get,

   f?(x) = (x + sec x) (1 - sec2x) + (x- tan x) (1 + tan x

...Read more

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New answer posted•a year ago
S
a year ago

71. Given, f (x) = x1+tanx

=x1+sinxcosx

=cosx·xcosx+sinx

So,f′(x)=(cosx+sinx)ddx(x⋅cosx)−(x⋅cosx)ddx(cosx+sinx)(cosx+sinx)2

 

=(cosx+sinx)(−xsinx+cosx)−(−x·sinxcosx+xcos2x)(cosx+sinx)2

=−xcosxsinx+cos2x−xsin2x+sinxcosx+xsinxcosx−xcos2x(cosx+sinx)2

=cos2x+sinxcosx−x(sin2x+cos2x)(cosx+csinx)2

Dividing numerator and denominator by cos2x we get,

=cos2xcos2x+sinxcosx·cosxcosx−x1cos2x(cosxcosx+sinxcosx)2

=1+tanx−xsec2x(1+tanx)2

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New answer posted•a year ago
S
a year ago

 70. Given, f (x) = x2cos (π4)sinx

=x2sinx×cosπ4

So, f? (x) =

 

=cosπ4 [2·xsinx−x2cosxsin2x]

=xcosπ4 [2sinx−xcosxsin2x].

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New answer posted•a year ago
S
a year ago

69. Given, f (x) = 4x+5sinx3x+7cosx

So,

f′(x)=(3x+7cosx)ddx(4x+5sinx)−(4x+5sinx)ddx(3x+7cosx)(3x+7cosx)2

=(3x+7cosx)(4x+5cosx)−(4x+5sinx)(3−7sinx)(3x+7cosx)2

=x+x  x+x+x−x+x  x− x+ 35 sin2x_                                                      ( 3x+7 cosx)2

=15xcosx+28cosx+28xsinx−15sinx+35(cos2x+sin2x)(3x+7cosx)2

=35+15xcosx+28cosx+28xsinx−15sinx(3x+7cosx)2

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