Let f(x) - 6 cos 6 cos° x f '(x) =

Q: Find f. f ''(?) = sin(?) + cos(?),    f(0) = 4,    f '(0) = 2

A: f''(x)= sin(x) + cos(x) , f(0)=4 ,f'(0) = 2

Q: Let f''(x) = 36 sin(6x) and suppose f'(0) 6 and f(0) = 4. Find f(5) a(;) =

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Q: f'(x) = Let f(x)= = I cos ³

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Q: If f(x) = cos a- 5 tanx, then %3D f'(x) = %3D f'(5) =

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Q: If f(r) 5 sin (2²), find f'(x). Find f'(0.5).

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Q: Let f'(x) = f'( 5 ) = f(x) = 3 sin x + 12 cos x

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Q: Find f'a) for f(x)= =Sinx 1- Cos X %3D

A: We have to find the derivative of the given function.

Q: Let f(x) = - 9 cos (sin(x*)). Then f'(x) = Preview

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Q: If f(x) = - f'(x) = f'(5) = 2 sin r 3 + cos x 3 then

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Q: Find f. f ''(?) = sin(?) + cos(?),    f(0) = 5,    f '(0) = 2 f(?) =

A: Given that        f"( θ) = sin(θ)+cos(θ)  The initial condition is f(0)=5 and f'(0)=2 We have to…

Q: Determine f " (x) a. f(x) = cot (4x) + csc (x) b. f(x) = sin2 (x) tan(x)

A: Given a. f(x)=cot4x+cscxb. f(x)=sin2xtanx

Q: If f(x) = 4x sin x cos x, find f'(x)= %3D

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Q: cos(4x) cos(2x) sin(4x) sec(2x) 12. Verify: - sin(2x)

A: cos(2x ) =cos2(x)-sin2(x) = 2cos2(x)-1 =1-2sin2(x) sin(2x) = 2sin(x) cos(x)

Q: Find f'(x) where f(x) sin(cos(4x2 +1))

A: Consider the given function

Q: f(x) = 8 cos x 5 tan x.

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Q: Let f(x) = (7 sin x + 2 cos x)tan-lx f'(x) = %3D

A: topic - derivatives

Q: et f(x)=-11sin x+3 cos x f' (x)=?

A: f(x)=-11sin x+3 cos x this is given  we have to differentiate the function  dsinx/dx = cosx…

Q: Let f(x) tan-(cos(32)) f'(x) =

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Q: 2x2 tan x f(x) sec x

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Q: Find f. f ''(?) = sin ? + cos ?,    f(0) = 1,    f '(0) = 3

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