Calculus problems.... stuck (derivatives, chain rule)?

Question

i have few questions...

1. If f(x) = x / cot(x) , then f' ( Pi / 4 ) =?
( this problem... i jsut dont get it....)

2. F(x) = G[x + G(x)] , howd you find F'(x) of this??

3. What is the fourth derivative of F(x) = (2x-3) ^ 4
( im not sure if this is 24(2^3) or 24(2^4))

thx

Answer 1

1. First, note that d/dx (cot x) = -cosec^2 x.

f(x) = x / cot x, so by the quotient rule we have
f'(x) = [1(cot x) - x(-cosec^2 x)] / cot^2 x
Now cot (π/4) = 1 and cosec (π/4) = √2, so we get
f'(π/4) = (1 - π/4 (-2)) / 1
= 1 + π/2.

2. F(x) = G[x + G(x)]
F'(x) = G'[x + G(x)] . d/dx (x + G(x)) by the chain rule
= G'[x + G(x)] . (1 + G'(x))

3. F(x) = (2x-3)^4
F'(x) = 4(2x-3)^3 . 2
F''(x) = 4.3.(2x-3)^2 . 2^2
F'''(x) = 4.3.2.(2x-3) . 2^3
F''''(x) = 24 . (2) . 2^3 = 24 . 2^4

Answer 2

Initially, I had made a mistake in my first answer which I have corrected after seeing Scarlet Manuka's answer.

1) f '(x)
= [cotx * d/dx(x) - x * d/dx(cotx) ] / cot^2(x)
= [ cotx + xcosec^2(x) ] / cot^2(x)
f' ( Pi / 4 )
= [ cot(pi/4) + xcosec^2(pi/4) ] / cot^2 (pi/4)
= 1 + (pi)/4 * 2 = 1 + (pi)/2

2) F ' (x)
= G ' [x + G(x)] * d/dx [x + G(x)]
= G ' [x + G(x)] * [1 + G '(x)]

3) F '(x)
= 4(2x - 3)^3 * 2 = 8 (2x - 3)^3
F" (x)
= 24 (2x - 3)^2 * 2 = 48 (2x - 3)^2
F"' (x)
= 96 (2x - 3) * 2 = 192 (2x - 3)
F""(x) = 384

Answer 3

pass the a million/3 in front of (a million+7x) then minus a million (or 3/3) from a million/3. Then, take the by-made of a million+7x, it is 7. so the stairs could look like [(a million/3) (a million+7x)^((a million/3) - (3/3))] (a million+7x)' [ (a million/3) (a million+7x)^(-2/3) ] * (7) answer is (7/3) * (a million+7x)^ ( - 2/3) or you may write it 7 ------------------------- 3(a million+7x)^(2/3) desire this clarification enables.