Possible Mistakes in Calculus Problems

[physicszman]

Please double check my work !

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

 

[Jupiter]

13) Find the 3rd derivative

y = 8x^2 - 4x + 7

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


third derivative does exist - it's just 0
y'=16x-4
y''=16
y'''=0

 

[Nexus[Free-DC]]

Physiczman,

your answer to q14 contains a mistake.





f(x)=x^3 + 3/x
f'(x)=3x^2-3/(x^2)
f''(x)=6x + 6/(x^3)
f'''(x)=6 - 18/(x^4)

This because 3/x = 3x^(-1).

 

[physicszman]

thansk guys.


Yea i tend to leave out the part where u square the denominator in the quotient rule. Thanks for the head up!

 

[physicszman]

for #3) Can I do

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

?


for #5) Can i do

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

?

Im not sure about 10 and 12 it seems like something could be done or?


TRhanks again guys i really appreciate it

 

[matt grime]

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?

I'm sure your professor respects you just as much. (A mistake is a mistake is a mistake.) Sorry that you think you should get better marks even though you got it wrong; let's hope he (or she) doesn't make any silly mistakes when assigning your grade at the end of the course.

 

[physicszman]

well i think i was exaggerating a bit when I said he was a dickhead. Hes a great guy but very picky.

When did I say or even imply that I want better marks even though I am getting it wrong?