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Question

Differentiate each the following from first principles :

(i) x sin x

(ii) x+cos x

(iii) sin (2x3)

(iv) sin 2x

(v) sin xx

(vi) cos xx

(vii) x2 sin x

(viii) sin(3x+1)

(ix) sin x+cos x

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Solution

(i) We have,

f(x)=x sin x

f(x)=limh0 f(x+h)f(x)h

=limh0 (x+h)sin(x+h)x sin xh

=limh0 x(sin(x+h)sin x)h+sin(x+h) [sin csin d=2cos c+d2sincd2]

=limh0 x×2(x+h2)sinh2h+sin(x+h)

=2x×cos x×12+sin x=x×cos x+sin x=sin x+x cos x

(ii) We have,

f(x)=x cos x

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

=limh0(x+h)cos(x+h)x cos xh

=limh0x cos(x+h)h cos(x+h)x cos xh=limh0x{cos(x+h)cos x}h+cos(x+h)

=limh0x.2sin(xxh2)sin(x+h2)+cos (x+h) [ cos Acos B=2sin BA2sinB+A2]

=limh02x.sin(h2)sin(x+h2)+cos (x+h)=x sin x+cos x

(iii) We have,

f(x)=sin (2x3)

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

=limh0sin{2(x+h)3}sin(2x3)h

=limh02 cos (2x+2h3+2x3)2×sin(2x+2h32x+3)2h

[ sin Csin D=2 cos C+D2sinCD2]

=limh0 2 cos(2x3+h).sin h2 [ lim00sin θθ=1]

=2 cos (2x3)

(iv) We have,

f(x)=sin 2x

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

=limh0sin(x+h)sin 2xh

Multiplying numerator and Denominator by sin 2(x+h)+sin 2x

=limh0sin 2(x+h)sin 2xh×sin 2(x+h)+sin 2xsin 2(x+h)+sin 2x

=limh0sin(2x+2h)sin 2x(sin 2(x+h)+sin 2x)

=limh0sin(2x+2h)sin 2x(sin(2x+2h)+sin 2x) [ sin Csin D=2 cos C+D2 sinCD2]

=limh02 cos(2x+h)×sin hh×1sin(2x+2h)+sin 2x

=2 cos 2x2sin 2x=cos 2xsin 2x

(v) We have,

f(x)=sin xx

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

=limh0sin(x+h)x+hsin xxh=limh0x sin(x+h)(x+h)sin xxh(x+h)

=limh0x(sin x.cos h+cos x.sin h)x.sin xh.sin xxh(x+h)

[ sin(A+B)=sin A.cos B+cos A.sin B]

=limh0x.sin x(cos h1)xh(x+h)+x.cos x.sin h(x+h)xhh sin x(x+h)xh [ 1cos h=2sin2 h2]

=x sin xx(x+h)×2sin2h2h22×h4+x cos xx2sin xx2

h0h20 and limθ0sin θθ=1

=0+x cos xsin xx2=x cos xsin xx2

(vi) We have,

f(x)=cos xx

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

=limh0cos(x+h)x+hcos xxh

=limh0x.cos(x+h)(x+h)cos x(x+h)×h [ cos (A+B)=cos A.cos Bsin A.sin B]

=limh0x[cos x.cos hsin x.sin h]x.cos xh.cos x(x+h)x.h

=limh0x cos x(cos h1)(x+h)x.hx.cos x.sin h(x+h)×hh.cos x(x+h)×h

=limh0x cos x.2sin2h2(x+h)×h24×h24s.sin xx(x+h)cos xx(x+h)

=0x sin xx2cos xx2=x sin xcos xx2

(vii) We have,

f(x)=x2sin x

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

=limh0(x+h)2sin(x+h)x2sin xh

=limh0(x2+h2+2hx)(sin x.cos h+cos x.sin h)x2 sin xh

[ sin (A+B)=sin A.cos Bcos B.sin A]

=limh0x2sin x(cos h1)h+h(h+2x)sin .cos hh+(x+h)2cos xsin hh

=limh0x2sin x×2sin2h2(h2)2+h24+(h+2x)sin x.cos h+(x+h)2cos x

=0+(2x sin x+x2 cos x)

=2x sin x+x2 cos x

(viii) We have,

f(x)=sin(3x+1)

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

=limh0sin 3(x+h)+1sin(3x+1)h

=limh0sin(3x+3h)+1sin(3x+1)h×sin(3x+3h)+1+sin(3x+1)sin(3x+3h)+1+sin(3x+1)

=limh0sin(3x+3h+1)sin(3x+1)h(sin(3x+3h)+1+sin(3x+1))

=limh02 cos (3x+1+3h2)×sin3h23h2×32×1sin(3x+3h)1+sin(3x+1)

=3cos(3x+1)2sin(x+1) [limh0sin3h23h2=1]

(ix) We have,

f(x)=sin x+cos x

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

=limh0{sin(x+h)+cos(x+h)}sin x+cos xh

=limh0{sin(x+h)+cos(x+h)sin xcos x}h

=limh0{sin(x+h)sin x}+{cos(x+h)cos x}h

=limh0{2sin(x+hx2cos(x+h+x2}+{2sinx+h+x2sinx+hx2}h

sin Asin B=2sinAB2cosA+B2and cos Acos B=2sin A+B2sinAB2

=limh0sin h.cos2x+h22sin(x+h2sin h)h

=limh0sin hh{cosx+h2sin(x+h2)} [ limh0sin hh=1]

=cos xsin x


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