Quick math question (1 Viewer)

[sando]

Find the 1st and 2nd Derivative of:

sqrt(2-x)

i hav managed to find the 1st but i jsut cant seem to get the second
i have come really close
it must just be a tiny mistake or somthin

so any answers????

 

[insert-username]

d/dx(√[2-x])

= d/dx([2-x]^1/2)

= -1/2(2-x)^-1/2

d/dx(-1/2(2-x)^-1/2)

= -1/4(2-x)^-3/2

= -1/4√(2-x)³

Is that what you get?


I_F

 

[sando]

ahh... thanks for that

i made simple error

i had it written different which made it extremly hard for me to work out

so now i get it

so thanks

 

[sando]

NEW QUESTIONS:
!!! !!!!!!!!!

1. Given that f ' ' (X) = 5x ^4 , f ' (0) = 3 and f (-1) = 1 . Find f (2)


2. A half pipe is to be made from a rectangle piece of metal of length x m, perimeter is 30 m. Find the dimensions of the rectangle that will give the maximum surface area.

I wouldnt have a clue how to work these out

 

[Trev]

f"(x)=5x⁴
So f'(x)=x⁵+C¹
f'(0)=3
∴ f'(x)=x⁵+3
f(x) = x⁶/6+3x+C²
f(-1)=1
∴ f(x)=23/6
f(x)=x⁶/6+3x+23/6
∴ f(2) = 20.5

 

[Trev]

Perimeter of the rectangle would have side lengths 15-x, 15-x, x and x.
Surface area = x(15-x)
SA=15x-x²
Finding stationary points (max and min points for surface area) is the derivative of SA.
(SA)'=15-2x
(SA)'=0 for stat. points.
15-2x=0; x=7.5
To show it is a max point (it obviously is, but yeah, you have to show these things for full marks).
(SA)"=-2; second derivitive is less than 0 therefore the point is a maximum.
Therefore dimensions are 7.5*7.5*7.5*7.5.

The max area in situations like this always end up being squares.

 

[sando]

nice work trev