find the derivative of sin(x+1) using first principles

ncert cbse 11th limits and derivatives miscellaneous exercise

1
(iii)find the derivative of sin(x+1) using first principles

f(x) = sin(x+1)

f(x+h) = sin[(x+h) +1]

f(x+h) - f(x) = sin[(x+h) +1] - sin[x+1]

using trigonometry formula trigonometry identities

sinx - siny =2cos[(x+y)/2] sin [(x-y)/2]

f(x+h) - f(x) = sin[(x+h) +1] - sin[x+1]

=2cos{ [(x+h) +1 +x+1 ] /2 } sin{ [(x+h) +1 - (x+1) ] /2}

=2cos{(2x+h+2)/2}sin{h/2}

[f(x+h) - f(x)] / h =[2cos{(2x+h+2)/2}sin{h/2} / h

=cos{(2x+h+2)/2} * [ sin(h/2) / (h/2) ] --------------(1)

using lim [ (sinx)/x ] =1 as x -->0

taking lim as h --> 0 in equation(1)

f ' (x) = cos{(2x+0+2)/2} *[1]

f ' (x) = cos{x+1} cancelling off the 2

(iv) find the derivative of cos[x-(pi/8)] using first principles

f(x) =cos[x-(pi/8)]

f(x+h) =cos[x+h-(pi/8)]

f(x+h) - f(x) = cos[x+h-(pi/8)]  - cos[x-(pi/8)]

 using trigonometry formula trigonometry identities

cosx - cosy = -2sin[(x+y)/2] sin[(x-y)/2]

f(x+h) - f(x) = cos[x+h-(pi/8)]  - cos[x-(pi/8)]

=-2sin{ {[x+h-(pi/8)]+[x-(pi/8)]}/2 }sin{ {[x+h-(pi/8)] - [x-(pi/8)]}/2 }

= (-2)sin{ {2[x-(pi/8)]+h}/2 } sin(h/2)

[f(x+h) - f(x)] / h =[ -2sin{ {2[x-(pi/8)]+h}/2 } sin(h/2) ] / h

= -sin{ {2[x-(pi/8)]+h}/2 }* [ sin(h/2) / (h/2) ] --------------(1)

using lim [ (sinx)/x ] =1 as x -->0

taking lim as h --> 0 in equation(1)

f ' (x)  = - sin[x-(pi/8)]

13. limits and derivatives
miscellaneous exercise

1.

(i)  find the derivative of (-x) using first principles
solution

(ii)   find the derivative of [ (-x)^(-1) ] using first principles

solution

(iii)find the derivative of sin(x+1) using first principles

solution

(iv) find the derivative of cos[x-(pi/8)] using first principles
solution




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