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Question Number 152904 by DELETED last updated on 03/Sep/21

Answered by DELETED last updated on 03/Sep/21

1). f(x)=4cos x+5sin x              (dy/dx)   =−4 sin x+5 cos ×  2). f(x)=3 sin 2x − 5 cos x               (dy/dx)  =3×2 cos 2x +5 sin x                 =6 cos 2x + 5 sin x//  3). f(x)= tan (2x+1)         (dy/dx) = 2 sec^2 (2x+1)//  4). f(x)= ((sin x + cos x)/(sin x))                 = 1 + cot x          (dy/dx) = −cosec^2  x

1).f(x)=4cosx+5sinxdydx=4sinx+5cos×2).f(x)=3sin2x5cosxdydx=3×2cos2x+5sinx=6cos2x+5sinx//3).f(x)=tan(2x+1)dydx=2sec2(2x+1)//4).f(x)=sinx+cosxsinx=1+cotxdydx=cosec2x

Answered by DELETED last updated on 03/Sep/21

f(x)=4cos x+5sinx   (d/dx) =u′v+uv′=0.cos x+4(−sin x)         =−4sin x         u=4 →u′=0         v=cos x→v′=−sin x        m=5→m′=0        n=sin x→n′=cos x  (d/dx)=m′n+mn′       =0.sin x+5.cos x=5cos x  →=−4sin x+5cos x

f(x)=4cosx+5sinxddx=uv+uv=0.cosx+4(sinx)=4sinxu=4u=0v=cosxv=sinxm=5m=0n=sinxn=cosxddx=mn+mn=0.sinx+5.cosx=5cosx→=4sinx+5cosx

Answered by DELETED last updated on 03/Sep/21

2) f(x)=3sin 2x−5cos x       (d/dx)=3.cos 2x ×2             =6cos 2x−5(−sin x)            =6cos 2x+5sin x

2)f(x)=3sin2x5cosxddx=3.cos2x×2=6cos2x5(sinx)=6cos2x+5sinx

Answered by DELETED last updated on 03/Sep/21

3). f′(x)=sec^2 (2x+1)(2)                =2 sec^2 (2x+1)

3).f(x)=sec2(2x+1)(2)=2sec2(2x+1)

Answered by DELETED last updated on 03/Sep/21

4). f(x)=((sin x+cos x)/(sin x))=1+cot x        (d/dx)=0−cosec^2 x=−cosec^2  x

4).f(x)=sinx+cosxsinx=1+cotxddx=0cosec2x=cosec2x

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