calculus
posted by alia .
*find the slope of y= 4/x +6
at the point (8,2)
*evaluate the derivative at the point (pi,4)
y=[tan(5x^8)]^3
please help

just tel me how to setup

I will assume your equation is as you typed it, and not
y = 4/(x+6)
y = 4/x + 6
= 4x^1 + 6
dy/dx = 4x^2
= 4/x^2
at (8,2) , dy/dx = 4/64 = 1/16
equation:
y2 = (1/16)(x  8)
16y  32 = x + 8
x + 16y = 40
y = [tan(5x^8)]^3
dy/dx = 3[tan(5x^8)]^2 (sec(5x^8) (40x^7)
at (π/4)
use your calculator to find the messy answer for dy/dx 
4/64
why 64 not 8 
notice dy/dx = 4/(x^2)
= 4/(8^2) = 4/64 or 1/16 
just plug pi/4
right 
thnks Reiny.
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