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Question

# Differentiate each the following from first principles : (i) x sin x (ii) x+cos x (iii) sin (2x−3) (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)=limh→0 f(x+h)−f(x)h =limh→0 (x+h)sin(x+h)−x sin xh =limh→0 x(sin(x+h)−sin x)h+sin(x+h) [sin c−sin d=2cos c+d2sinc−d2] =limh→0 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)=limh→0f(x+h)−f(x)h =limh→0(x+h)cos(x+h)−x cos xh =limh→0x cos(x+h)h cos(x+h)−x cos xh=limh→0x{cos(x+h)−cos x}h+cos(x+h) =limh→0x.2sin(x−x−h2)sin(x+h2)+cos (x+h) [∵ cos A−cos B=2sin B−A2sinB+A2] =limh→02x.sin(−h2)sin(x+h2)+cos (x+h)=−x sin x+cos x (iii) We have, f(x)=sin (2x−3) ∵ f′(x)=limh→0f(x+h)−f(x)h =limh→0sin{2(x+h)−3}−sin(2x−3)h =limh→02 cos (2x+2h−3+2x−3)2×sin(2x+2h−3−2−x+3)2h [∵ sin C−sin D=2 cos C+D2sinC−D2] =limh→0 2 cos(2x−3+h).sin h2 [∵ lim0→0sin θθ=1] =2 cos (2x−3) (iv) We have, f(x)=√sin 2x ∵ f′(x)=limh→0f(x+h)−f(x)h =limh→0√sin(x+h)−√sin 2xh Multiplying numerator and Denominator by √sin 2(x+h)+√sin 2x =limh→0√sin 2(x+h)−√sin 2xh×√sin 2(x+h)+√sin 2x√sin 2(x+h)+√sin 2x =limh→0sin(2x+2h)−sin 2x(√sin 2(x+h)+√sin 2x) =limh→0sin(2x+2h)−sin 2x(√sin(2x+2h)+√sin 2x) [∵ sin C−sin D=2 cos C+D2 sinC−D2] =limh→02 cos(2x+h)×sin hh×1√sin(2x+2h)+√sin 2x =2 cos 2x2√sin 2x=cos 2x√sin 2x (v) We have, f(x)=sin xx ∵ f′(x)=limh→0f(x+h)−f(x)h =limh→0sin(x+h)x+h−sin xxh=limh→0x sin(x+h)−(x+h)sin xxh(x+h) =limh→0x(sin x.cos h+cos x.sin h)−x.sin x−h.sin xxh(x+h) [∵ sin(A+B)=sin A.cos B+cos A.sin B] =limh→0x.sin x(cos h−1)xh(x+h)+x.cos x.sin h(x+h)xh−h sin x(x+h)xh [∵ 1−cos h=2sin2 h2] =−x sin xx(x+h)×2sin2h2h22×h4+x cos xx2−sin xx2 ∵ h→0⇒h2→0 and limθ→0sin θθ=1 =0+x cos x−sin xx2=x cos x−sin xx2 (vi) We have, f(x)=−cos xx ∵ f′(x)=limh→0f(x+h)−f(x)h =limh→0cos(x+h)x+h−cos xxh =limh→0x.cos(x+h)−(x+h)cos x(x+h)×h [∵ cos (A+B)=cos A.cos B−sin A.sin B] =limh→0x[cos x.cos h−sin x.sin h]−x.cos x−h.cos x(x+h)x.h =limh→0x cos x(cos h−1)(x+h)x.h−x.cos x.sin h(x+h)×h−h.cos x(x+h)×h =limh→0−x cos x.2sin2h2(x+h)×h24×h24−s.sin xx(x+h)−cos xx(x+h) =0−x sin xx2−cos xx2=−x sin x−cos xx2 (vii) We have, f(x)=x2sin x ∵ f′(x)=limh→0f(x+h)−f(x)h =limh→0(x+h)2sin(x+h)−x2sin xh =limh→0(x2+h2+2hx)(sin x.cos h+cos x.sin h)−x2 sin xh [∵ sin (A+B)=sin A.cos Bcos B.sin A] =limh→0x2sin x(cos h−1)h+h(h+2x)sin .cos hh+(x+h)2cos xsin hh =limh→0−x2sin 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)=limh→0f(x+h)=f(x)h =limh→0√sin 3(x+h)+1−√sin(3x+1)h =limh→0√sin(3x+3h)+1−√sin(3x+1)h×√sin(3x+3h)+1+√sin(3x+1)√sin(3x+3h)+1+√sin(3x+1) =limh→0sin(3x+3h+1)−sin(3x+1)h(√sin(3x+3h)+1+√sin(3x+1)) =limh→02 cos (3x+1+3h2)×sin3h23h2×32×1√sin(3x+3h)1+√sin(3x+1) =3cos(3x+1)2√sin(x+1) [limh→0sin3h23h2=1] (ix) We have, f(x)=sin x+cos x ∵ f′(x)=limh→0f(x+h)−f(x)h =limh→0{sin(x+h)+cos(x+h)}−sin x+cos xh =limh→0{sin(x+h)+cos(x+h)−sin x−cos x}h =limh→0{sin(x+h)−sin x}+{cos(x+h)−cos x}h =limh→0{2sin(x+h−x2cos(x+h+x2}+{−2sinx+h+x2sinx+h−x2}h ⎡⎣∵ sin A−sin B=2sinA−B2cosA+B2and cos A−cos B=2sin A+B2sinA−B2⎤⎦ =limh→0sin h.cos2x+h2−2sin(x+h2sin h)h =limh→0sin hh{cosx+h2−sin(x+h2)} [∵ limh→0sin hh=1] =cos x−sin x

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