# calculus

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*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

• calculus -

just tel me how to setup

• calculus -

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:
y-2 = (-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)

• calculus -

-4/64
why 64 not 8

• calculus -

notice dy/dx = -4/(x^2)

= -4/(8^2) = -4/64 or -1/16

• calculus -

just plug pi/4
right

• calculus -

thnks Reiny.

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