Question

Find the derivative of the following functions from first principle:
(i) −x (ii) (-x)-1 (iii) sin (x + 1) (iv) cos (x – π/ 8 )

Answer 1

(i) Let f(x) = -x. Accordingly, f(x+h)=-(x+h)
By first principle,
f'(x)=lim h→0 f(x+h)-f(x)/h
=lim h→0 -(x+h)-(-x)/h
=lim h→0 -x-h+x/h
=lim h→0 -h/h
=lim h→0(-1)=-1

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(ii) Let f(x)=(-x)-1 = 1/-x = -1/x Accordingly, f(x+h)=-1/(x+h)
By first principle,
f'(x)=lim h→0 f(x+h)-f(x)/h 
=lim h→0 1/h[-1/x+h - (-1/x)]
=lim h→01/h[-1/x+h + 1/x]
=lim h→01/h[-x+(x+h)/x(x+h)]
=lim h→01/h[-x+x+h/x(x+h)]
=lim h→01/h[h/x(x+h)]
=lim h→0 1/x(x+h)
= 1/x.x - 1/x2

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(iii) Let f(x) = sin (x + 1). Accordingly, f(x+h)=sin(x+h +1)
By first principle,
f'(x)=lim h→0 f(x+h)-f(x)/h 
=lim h→0 1/h[sin(x+h+1) -sin(x+1)]
=lim h→0 1/h[2cos(x+h+1+x+1/2) sin(x+h+1-x-1)]
=lim h→0 1/h[ 2cos (2x + h+2/2) sin(h/2)]
=lim h→0 1/h[ 2cos (2x + h+2/2) sin(h/2)/(h/2)]
=lim h→0 cos (2x + h+2/2).lim h/2→0 sin(h/2)/(h/2)] [As h→0 ⇒h/2 →0]
=cos(2x+0+2/2).1 [lim h→0 sinx/x =1]
=cos(x+1)

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(iv) letf(x) cos(x-π/8). Accordingly, f(x+h)=cos(x+h -π/8)
By first principle,
f'(x)=lim h→0 f(x+h)-f(x)/h 
=lim h→01/h[cos(x+h-π/8)- cos(x-π/8)]
=lim h→01/h[-2sin(x+h-π/8+x-π/8)/2 sin(x+h-π/8-x+π/8 /2)]
=lim h→01/h[-2sin(2x+h-π/4 /2) sin h/2]
=lim h→0[-sin(2x+h-π/4 /2) sin (h/2)/(h/2)]
=lim h→0[-sin(2x+h-π/4 /2)].lim h/2→0 sin (h/2)/(h/2) [As h→0 ⇒h/2 →0]
=-sin(2x+0-π/4 /2) .1
=-sin(x-π/8)