Some other derivative help, I don't understand this. >.<?

Question

I'm not understanding this derivative crap at all, why did they ever make it?

a. Use the definition of the derivative to find f'(x) if f(x) = 4/x when x = 8

b. Find the left and right hand derivatives of f(x) at (1,4) if

f(x) = (x^2 + 3x, x >1) and (x^3 + x^2 + x + 1, x =< 1)


c. Find the equation of a line normal to f(x) if f(-2) = 6 and f'(-2) = 2/3


Thanks guys for all your help!

Answer 1

a) f'(x)=4x^-1=-4x^-1-1=-4x^-2=-4/x^2, now plug in initial condition x=8, -4/x^2=-4/8^2=-4/64=-1/16

b)f'(x)=2x+3>1, 2x>-2, therefore x>-1
f'(x)=3x^2+2x+1=<1
=3x^2+2x=<0

i forgot how to do c though, basically (-2,6) and (-2,2/3) and just plug those into f(x) and f'(x) and then find the equation

Answer 2

c. Recall that if the slope of a line is m, the slope of any line perpendicular to that line is -1/m. Since f'(-2) = 2/3 is the slope of the tangent line at (-2,6), the slope of the normal line at (-2,6) is -3/2. Thus, an equation for the normal line is

y-6 = (-3/2)(x-(-2))

BTW, the computer you're using right now would be impossible if we didn't have "this derivative crap".