Differentiation!?

Question

Can anyone help me? I have a maths textbook but like my classbook it never tells me the method to do the very hard questions. Like all the differentiation questions are f(x)= x^3(x^4+x^6), never a question or example that tells me how to differentiate for example the one below. Your help will be greatly appreciated! Thanks!

f(x) =(x+2)^3 how do you differentiate the equation when it is to ^3?

f(x) = x ^ (3/2) - 8 x ^ (-1/2)
Evaluate f(3) giving the answer in its simplest form with a rational denominator?

How do i evaluate (5 and 4/9) ^ (-1/2)
and find the value of x where (1+x) / x = sqrt 3 giving answer = a+b sqrt 3


Straight line l is perpendicular to line with equation 3x-y=4and passes through the point with co-ords (3,0)
How do i find l? When it is perpendicular do i just put 1 over?

Also how do i find the co-ordinates of the point where l interesect with line y= 2x-13?

Answer 1

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

NB The '1' comes about by differentiating the term inside the brackets. If g(x) = x + 2 then g'(x) = 1
f(x) = x^(3/2) - 8x^(-1/2)
f'(x) = [3x^(1/2)]/2 - [-1x8x^-(3/2)] / 2

f(3) = 3^(3/2) + 8(3)^(-1/2)
f(3) = (sq rt 3)^3 - 8/sq rt 3
f(3) = [(sq rt3)^4 - 8] / sq rt 3
f(3) = [9 - 8] / sq rt 3
f(3) = 1 / sq rt 3
f(3) = sq rt 3 / 3

(5 4/9)^ -1/2
convert 5 4/9 to improper fraction
5 4/9 = 49/9
Hence (49/9)^ - 1/2 = (9/49)^1/2
(9/49)^1/2 = sq rt (9/49)
sq rt (9/49) = sq rt 9 / sq rt 49
= 3 /7

(1 + x)/x = sq r 3
1/x + 1 = sq rt 3
1/x = -1 + sq rt 3
x = 1/( - 1 + sq rt 3) Apply the conjugate (-1 - sq rt 3)
x = 1 ( -1 -sq rt3) /(-1 + sq rt 3)(-1 - sq rt 3)
x = -1 - sq rt 3 / 1 - 3
x = -1 - sq rt 3 / -2
x = -1/-2 - (sq rt 3)/-2
x = 1/2 + sq rt 3 /2
hence 'a' = 1/2 and 'b' = 1/2

3x - y = 4
y = 3x - 4
the slope is '3'
To find the perpendicular line apply the rule MM' = -1 were M & M' are the two slopes.
3 M' = -1
M' = -1/3
y = -1x/3 + c at the point (3,0)
0 = -1(3)/3 + c
0 = -1 + c
c = 1
hence y = -x/3 + 1
or 3y + x = 3

The intersection of the two lines y = 2x -13 & 3y + x = 3
Resolve simultaneously.

2x - y = 13
x + 3y = 3 (Multtiply by -2 to eliminate 'x')

2x - y = 13
-2x - 6y = -6 (& add)

-7y = 7
y = -1

substitute y= -1 into either equation to find 'x'
x + 3(-1) = 3
x -3 = 3
x = 6
Hence co-ords of interssection are (6, -1)

Answer 2

d/dx (x+2)^3 = 3(x+2)^2 d/dx(x+2) = 3(x+2)^2 *1 = 3(x+2)^2

f(3) = 3^3/2 -8 3^-1/2 = 3sqrt(3) -8/sqrt(3)
multiply by sqrt(3)/sqrt(3) [=1]
=9 - 8sqrt(3)/3 = (27-8sqrt(3))/3

(5 4/9)^-1/2 = (49/9)^-1/2 = (9/49)^1/2 = 3/7

3x-y = 4
y = 3x-4
slope 3
slope of perpendicular = -1/3
y = -1/3 x + b
plug in (3,0)
0 = -1 + b
b = 1
y = -1/3x + 1

y = -1/3x + 1
y = 2x-13
-1/3 x + 1 = 2x-13
5/3 x = 14
x = 42/5