Limits and Derivatives

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alok kumar singh

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48. Given, f (x) = ax+bcx+d

So, f (x) =  (cx+d)ddx (ax+b) (ax+b)ddx (cx+d) ( (x+d)2

= (cx+d)a (ax+b)c (cx+d)2

=acx+adacxcb (cx+d)2

=adbc (cx+d)2

New question posted

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New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

47. Given, f (x) = (ax + b) (cx + d)2

So, f?(x) = (ax +b) ddx(cx+d)2+(cx+d)2ddx(ax+b)

=(ax+b)ddx[(c+d)(cx+d)]+(cx+d)2a

=(ax+b)[(c+d)ddx(x+d)+(c+d)ddx(c+d)]+a(cx+d)2

=(ax+b)[c(cx+d)+c(cx+d)]+a(cx+d)2

=(ax+b)·2·c(cx+d)+a(cx+d)2

New answer posted

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alok kumar singh

Contributor-Level 10

44.
= l i m h 0 x h + x h
= l i m h 0 h h
= l i m h 0 1

= - 1.

(ii) Given, f(x) = (-x)-1

by first principle,

f(x) l i m h 0 [ ] ( x + h ) 1 ( x ) 1 h

(iii) Given, f(x) = sin(x + 1)

By first principle,

f'(x) = limh0f(x+h)f(x)h

=limh0sin(x+h+1)sin(x+1)h

=limh02h·cos(x+h+1+x+12)·sin(x+h+1(x+1)2)

=limh02hcos(2x+2+h2)sin(h2).

=limh0cos(2x+2+h2)limh0sinh2h2

=cos(2x+2+02)*1.

= cos (x + 1)

(iv) Given, f(x) = cos (xπ8)

By first principle,

f(x) = limh0f(x+h)f(x)h

=limh0cos(x+hπ8)cos(xπ8)h

 

New answer posted

4 months ago

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alok kumar singh

Contributor-Level 10

. (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)+5lim

New answer posted

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alok kumar singh

Contributor-Level 10

(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

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

42. Given, f(x) = cos x

f(x+h)=cos(x+h)

By first principle,

f(x)=limxhf(x+h)f(x)h

=limxh1h[cos(x+h)cosx]

=limxh1h[2sin(x+h+x2)sin(x+hx2)]

=limxh1h[2sin(2x+h2)sin(h2)]

=limxh1h[2sin(2x+h2)sin(h2)]

=limxhsin(2x+h2)*limxhsinh/2h/2

=sin(2x+02)*1=sinx.

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

41. (i) f(x)=2x34

f(x)=ddx(2x34)

=2dxdx0

=2.

(ii) Given, f(x)= (5x3+3x1)(x1)

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

=(5x3+3x1).1+(x1)(15x2+3)

5x3+3x1+i5x3+3x15x23

=20x315x2+6x4.

(iii) Given, f(x) = x3(5+3x)

So, f(x)=x3ddx(5+3x)+(5+3x)dx3dx=x33+(5+3x)(3)x4

=3x315x49x3

=15x46x3

=3x4(5+2x).

(iv) Given, f(x)= x5(36x9).

f(x)=x5ddx(36x9)+(36x9)ddxx5.

=x5(6x9x10)+(36x9)5x4

x5(54x10)+15x430x5

=54x530x5+15x4

=24x5+15x4

(v) Given, f(x)= x4(34x5).

So, f(x)=x4ddx(34x5)+(34x5)ddxx4

=x4(4x5*x6)+(34x5)(4x5)

=20x1012x5+16x10.

=36x1012x5.

=36x1012x5.

(vi) Given, f(x)= 2x+1x23x1

So, f(x)=ddx(2x+1)ddx(x23x1)

=(x+1)ddx2ddx(x+1)(x+1)2(3x1)dx2dxx2ddx(3x1)(3x1)2

=2(x+1)22x(3x1)3x2(3x1)2

=2(x+1)26x22x3x2(3x1)2

=2(x+1)23x22x(3x1)2

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

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

40. Given, f(x)= xnanxa.

So, f(x)=(xa)ddx(xna3)ddx(xa)(xnan)(xa)2=(xa)nxn1(xnan)(xa)2

=(xa)nxn1(xnan)(xa)2

=nxn1xnaxn1xn+an(xa)2

=nxnxxnaxn1+an(xa)2

New answer posted

4 months ago

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A
alok kumar singh

Contributor-Level 10

(c) Given, f(x)=(x-a)(x-b)

where a and b are constants.

So,

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

=(x-a)+(x-b)

= 2x– a- b. 

(ii) Given f(x)= (ax2+b)2. where ab are constant

So, f(x)=ddx(ax2+b)2

=ddx(a2x4+b2+2ax2b)

=ddxa2x4+ddxb2+ddx2ax2b

4a2x3+0+4axb

4ax(ax2+b).

(iii) Given, f(x)= x9xb where a and bare constants

So, f(x)=(xb)ddx(xa)(xa)ddx(xb)(xb)2

=(xb)(xa)(xb)2

=ab(xb)2

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