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Hi, I basically did out all the poroblems. I just need someone to double check whether I made any mistakes in the answers or procedures. (My prof is a real dickhead and marks down for the smallest mistake) Also can I simplify any answers, just like obvious stuff?

Thanks for the help!!!

1)

f(x) = x / (400-x)

f'(x) = (400-x)(1) - (-1)(x) / (400-x)^2

f'(x) = (400-x+x) / (400-x)^2

f'(x) = 400 / (400-x)^2

2)

g(x) = (2x+3)/(x-5)

g'(x) = (x-5)(2)-(1)(2x+3) / (x-5)^2

g'(x) = (2x-10-2x-3) / (x-5)^2

g'(x) = -13 / (x-5)^2

3)

y = (3x-4)/(x^3+1)

y' = (x^3+1)(3x)-(3x^2)(3x-4) / (x^3+1)^2

y' = [(3x^4+3x)-9x^3+12x^2] / (x^3+1)^2

y' = (3x^4-9x^3+12x^2+3x) / (x^3+1)^2

4)

f(x) = (3x^2+2x) / x^5

f'(x) = (x^5)(6x+2)-(5x^4)(3x^2+2x) / (x^5)^2

f'(x) = (6x^6+2x^5-15x^6-10x^5) / (x^5)^2

f'(x) = (-9x^6-8x^5) / (x^5)^2

5)

h(x) = sqrt(3+2x)

h(x) = (3+2x)^(1/2)

h'(x) = 1/2(3+2x)^(-1/2)(2)

6)

y = (x^4-4x^2+x)^-5

y' = -5(x^4-4x^2+x)^-6(4x^3-8x+1)

7)

y = (x)sqrt(3x+4)

y' = (x)1/2(3x+4)^(-1/2)(3)(sqrt(3x+4))

8)

f(x) = x^3 + (100-x)^2

f'(x) = 3x^2 + 2(100-x)(-1)

9)

y = 1 / (x+2)^2

y = (1)(x+2)^-2

y'= -2(x+2)^-3(1)

y'= 1/(-2(x+2)^3)

10)

h(x) = 1 + sqrt(x) / (x^5+3)

h'(x) = (x^5+3)(1+1/2(x)^(-1/2))-5x^4(1 + sqrt(x)) / (x^5+3)^2

11)

g(x) = (x-4)^8 * (x+3)^9

g'(x) = (x-4)^8 * 9(x+3)^8 + 8(x-4)^7 * (x+3)^9

12)

f(x) = sqrt((3+x)/(2-x))

f'(x) = 1/2(((3+x)/(2-x))^-1/2 * (((2-x) + (3+x)) / (2-x)^2)

13) Find the 3rd derivative

y = 8x^2 - 4x + 7

3rd derivative does not exist because highest power is 2.

14) Find the 3rd derivative

f(x) = x^3 + 3/x

f'(x) = 3x^2 + (-3)

f'(x) = 6x

f'(x) = 6

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# Please double check my work

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