Please Show Me Step By Step How To Find The...

Anonymous

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Please show me step by step how to find the derivativeof a.) cos(3x+sinx) b.) 94x-5+3log9(2x+1) Thanks for your help- I will give high ratings

 

Answer No.1

DedicatedGuy28

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A.) dy/dx(cos(3x+sinx))= -sinx(3x+(sinx))dy/dx(3x+sinx) beacuse in this condition u hv also to take the derivative ofthe x. means cos(x).after taking derivative of cosx u should takederivative of x. =-sinx(3x+(sinx)).3.(cosx) =-3cosxsinx(3x+(sinx)). solution b.) dy/dx(94x-5+3log9(2x+1))=log94x-5.dy/dx(4x-5)+dy/dx(3log9(2x+1)) =log94x-5.4+dy/dx(3log32(2x+1)) =4 log 94x-5+6(1/(2x+1)).dy/dx(2x+1) = 4 log 94x-5+2.6/(2x+1)

Answer No.2

Anonymous

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(A). F(X)=COS[3X+SINX] F'(X)=-SIN[3X+SINX] [3+COSX] FORMULAS: (1) D/DX[SINU]=(COSU)U' (2) D/DX[COSU]= (-SINU)U'

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