Need help with this calc derivative problem! PLEASE?!
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Need help with this calc derivative problem! PLEASE?!

[From: ] [author: ] [Date: 11-12-15] [Hit: ]
In addition,First,The derivative of a sum or difference is the sum or difference of the derivatives.The derivative of a constant is zero.The derivative of a constant is zero.The derivative of a constant is zero.......
http://tinypic.com/r/9htbgw/5

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It is VERY hard to read. Here's what I see:

g(x) = f(x) - 2
h(x) = 2f(x)
r(x) = f(-3x)
s(x) = f(x + 2)

In addition, you know

f '(-2) = 4
f '(-1) = 2/3
f '(0) = -1/3
f '(1) = -1
f '(2) = -2
f '(3) = -4

First, the derivatives:

g(x) = f(x) - 2 ⇒ g'(x) = f '(x)
h(x) = 2f(x) ⇒ h'(x) = 2f '(x)
r(x) = f(-3x) ⇒ r '(x) = -3f '(-3x)
s(x) = f(x + 2) ⇒ s'(x) = f '(x + 2)

using various elementary differentiation rules:

for g'
The derivative of a sum or difference is the sum or difference of the derivatives.
The derivative of a constant is zero.

for h'
Product rule
The derivative of a constant is zero.

for r '
Chain rule
The derivative of a constant is zero.

for s'
Chain rule
The derivative of a sum or difference is the sum or difference of the derivatives.
The derivative of a constant is zero.

Since g' = f ', you need only copy the f ' values for each x into the corresponding spot in the g' row.

h'(x) = 2f '(x), so double the f ' value for each x and put this into the corresponding spot in the h' row.

Now for the fun part.

r '(x) = -3f '(-3x), so

r '(-2) = -3f '(6)
r '(-1) = -3f '(3)
r '(0) = -3f '(0)
r '(1) = -3f '(-3)
r '(2) = -3f '(-6)
r '(3) = -3f '(-9)

Of the f ' values needed to fully fill in the r ' row, the table tells us only f '(3) and f '(0), so you can fill in only r '(-1) and r '(0). r '(-1) = -3f '(3) = -3(-4) = 12 and r '(0) = -3f '(0) = -3(-1/3) = 1.

Finally, s'(x) = f '(x + 2). Write down s'(-2), s'(-1), etc. as I've done for r ', and see which of the s' values you can fill in. (it will be all but for x = 2 and x = 3).
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