If f(x) = (cos x)(e^sin x ), then, f '(4.0) = 0.08. Is this correct?

Answer 1

NO

f'(x)=(-sinx)(e^cosx)
f'(4.0)=(-.07)(2.7)= -1.89

Answer 2

f(x) = (cos x)(e^sin x)

f '(x) = (cos x)(e^sin x)(cos x) + (e^sin x)(- sin x)
= (cos^2 x)(e^sin x) - (sin x)(e^sin x)
= (e^sin x)[ cos^2(x) - sin x ]

f '(4.0) = (e^sin 4.0)[ cos^2(4.0) - sin 4.0 ]
= 2.71828183^(-0.75680250) [ 0.42724998 - (-0.75680250) ]
= 0.46916419 [ 1.18405248 ]
= 0.55551502
= 0.56

I checked that answer by graphing both f(x) and f '(x),
so I think it must be correct, so I guess 0.08 is incorrect.

Answer 3

Say f(x) =y=(cosx)(e^sinx)
Lny= lncosx+sinx
1/y dy/dx= 1/cosx.(-sinx) + cos x
f'(x) = (cosx)(e^sinx) (-tanx + cosx)
f'(4) = (cos4)(e^sin4)( cos4-tan4)
= 0.997* 1.07*(0.997-0.07)
= 0.99
So 0.08 is not correct Ans.