f(x) = (-x² + x)ln(x³ + 2x) f'(x) = ? %3D

Q: f(x)=sqrt(5x^2+5x+8) f′(x)=    f'(5)=

A:

Q: Find f'(1) for f(x) = ln(3x² – 6x – 2) rounded to 3 decimal places. f'(1) = |6-6x/3x^2-6x-2

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Q: ∫(3/2 sqrt(x )-4/x^2 ) dx ]

A:

Q: Differentiate f(x) = (3x)n() for r > 0. %3D

A: To differentiate (variable)(variable) ,  we need to take logarithm both the side then to…

Q: f(x) = 16x f'(x) = – 4x 4 f'(x) = 64x 4 f'(x) = 12x4 4 f'(x) = 4x

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Q: f(x)=9/(5x^2+4x+1)^2 find f'(x)

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Q: h(x)= 8/x-2/x^3+1/x^4 Find H'(x)

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Q: F(x)= 3e^(4x^2+4x+4) Find F'(0) round answer to 4 decimal places

A: Given,           F(x)=3e4x2+4x+4

Q: ln(x+(x2 −16)^(1/2)) Find the domain and the derivative of the function.

A: Given, function is

Q: Find f'(x) and f"(x). f(x) x2 - 8 f"(x) f"(x)

A:

Q: f(x)=ln(x^5) f′(x)= f′(e^3)=

A: Given query is to find indicated derivative.

Q: Let (v 6x - 8 f(x) = ln %3D 5x + 4 f'(x) =

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Q: n'(x) = Find the derivative of n(x) = (13x + 4) (x³ + √√x)

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Q: if f(x)=(4x+8)^-4 find f'(x) and f'(3).

A: Given:- f(x) = (4x+8)-4 To find:- f'(x) and f'(3)

Q: f(x)=ln[x^7(x+2)^8(x^2+8)^2] f'(x)=?

A: Given that: f(x)=lnx7(x+2)8(x2+8)2

Q: f(x)=(1/x^3)^1/2 f’(x)=?

A: Given function is,

Q: 2hv3 1+ In-1 I,c²

A: Let's find.

Q: Find f'(x) for f(x) = %3D (x-4)² f'(x) = %3D

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Q: F(x)=e^2x (3x^2-x)

A: see 2nd step

Q: f(x)=e^4x,x=3pi what is the value

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Q: d/dx (x^6 + ln(x) + 21)

A: given ddxx6+ln (x) +21

Q: f(x)=−6indexroot3√x−1/x^8 find f'(x)=

A: Given,           fx=-6x3-1x8

Q: 二十三 _{{x} = {\}\

A: Assume the bracket part as another variable and then use Chain Rule. And use basic identities of…

Q: f(x)=3x^1/5(x^3-9) f'(3)=?

A: The objective of the question is find the derivative at x=3 . That is f'(3).

Q: If f(x)=(4x)^(ln x) how would you find f'(3)?

A: Obtain the derivative of the function as follows.

Q: f(x) = (2x2 – 5x)ln(x² – x) f'(x) = ?

A: Product Rule ddx(h(x)g(x))=ddx(h(x))g(x)+h(x)ddx(g(x)) Chain Rule ddx(h(g(x)))=ddhu(h(u))ddx(g(x))…

Q: f(x)=8^(x^2+4x) f'(1)=?

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Q: X Je* tan-¹ (e^x) dx е

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Q: f'(x)=3x^2-6x-3. =3(x-1)^2-6. For what values of x is f'(x)=0?

A: f'(x)=03(x-1)2-6=03(x-1)2=6 (x-1)2=63  (x-1)2=2 (x-1)=±2