Let f(x) - 6 cos 6 cos° x f '(x) =

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: Let f(x) = 9 cos(sin(z°)). %3D Then f'(x),

A: Using the chain rule of differentiation,

Q: Let f(x) = 12 sin x + 11 cos x f'(x) = f'(꽃) =D

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Q: sin (Vx).cos (Vx) dx = ?

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Q: f'(x) = Let f(x)= = I cos ³

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Q: 3x2 tan x If f(x) sec x

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Q: Find the critical point of the function f(x) = 3x² – 12x Answer:

A: We will find out all the required values.

Q: Let f(x) 4 cos x + 2 tan x f' (æ) = r(금)- f'

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Q: Let .f(x) = - 4 cos? x f'(x) =

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Q: If f(x) f'(x) = = COs x 4 tan x, then f'(4) = Check Answer

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Q: 1 Let f(x) = (5 sin x + 8 cos a)tan f'(x) =

A: NOTE: Refresh your page if you can't see any equations. . take derivative use the product rule for…

Q: f(x) = 2 sin(cos z)

A: Given,           y = 2 sin(cos x)

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

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Q: Let f(x) = sin(x)x-2 Find f'(x).

A: Differentiation Question

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

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

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: 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: f(x) = 2 tan x Ꮖ

A: f(x)=2tan⁡xx

Q: Let f(x) = – 8 cos(cos (a®)). Then f'(x) =

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Q: If f(x) f'(æ) = cos x – 7 tan x, then

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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: Let h(x) = tan(3x + 7). Then n'(5) is and h'"(5) is

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Q: x sin x f(x) = = -9e | f'(x) =

A: Use product and chain rule of derivative

Q: Let f(x)=(6sinx+2cosx)tan−1x f'(x) =

A: Consider the function

Q: f(x) = 7 sinx+ 5 cos x

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Q: df. ƒ "''(x) = cos(x), f(0) = 8, f'(0) = 7, ƒ "(0) = 8

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