If f(x) = (cos x)(e^sin x ), then f '(2.0) =?

Question

possible solutions

a. -2.69
b. -1.83
c. 0.08
d. 0.56
e. 0.86
f. 0.97
g. none of these

Answer 1

Just lots of chain rule and product rule applications.

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

Plug in 2 and chug your way to an answer.

Answer 2

In order to do this question, you have to use the product rule and the chain rule.

Reminder of product rule: derivative of the first times the second plus the first times the derivative of the second.

Reminder of the chain rule: take the derivative as it were just x, but take the derivative of the inside and multiply it.

f'(x) = (-sin x) (e^(sin(x))) + (cos x) (e^(sinx))(cosx)
f'(x) = e^(sin(x)) [ -1 + [cos(x)]^2 ]

Using a nifty trig identity, you'll learn that [ -1 + [cos(x)]^2 ] is equal to - [sin(x)]^2, so

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

From this point, all you have to do is plug in 2 for x.

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

Which you just approximate with a calculator.

Answer 3

if u have a good calculator (TI-83 might work, TI-89 will definitely) then u should just plug it in. f'(x)=(cos x)^2 * (e^sin x) + (e^sin x)(-sin x) by the chain rule. Plugging in results in an answer of -1.83.

Answer 4

f'(x) = (-sin x)(e^sin x) + (cos x)(e^sin x)(cos x). (This is by the Product Rule; note the derivative of (e^sin x) is (e^(sin x))(cos x) by the Chain Rule.)

f'(2.0) = (-sin 2.0)(e^(sin 2.0)) + (cos 2.0)(e^(sin 2.0))(cos 2.0) = about -1.83.

So the answer is (b).

Answer 5

f'(x) = -sin(x)(e^sin x) + cos(x)(cos(x)(e^sin x)) by the chain and product rules.
Now plugging in 2 for x...
the answer is b, -1.83

Answer 6

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

f'(x) = (-sen x)(e^sin x) + (cos x)(cos x*e^sin x)
f'(x) = (-sen x)(e^sin x) + (cos²x*e^sin x)
f'(x) = (cos²x-sen x)(e^sin x)

=>
f'(2) = (cos²2-sen 2)(e^sin 2) ............... NOT IN DEGREES!!!

f'(2) = -1,8274732235788288210478600385165

Answer: letter (b)

₢