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October 21, 2014

October 21, 2014

Posted by **alia** on Sunday, December 16, 2012 at 9:14pm.

at the point (8,2)

*evaluate the derivative at the point (pi,4)

y=[tan(5x^8)]^3

please help

- calculus -
**alia**, Sunday, December 16, 2012 at 10:40pmjust tel me how to setup

- calculus -
**Reiny**, Sunday, December 16, 2012 at 10:48pmI 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)

use your calculator to find the messy answer for dy/dx

- calculus -
**alia**, Sunday, December 16, 2012 at 10:56pm-4/64

why 64 not 8

- calculus -
**Reiny**, Sunday, December 16, 2012 at 11:06pmnotice dy/dx = -4/(x^2)

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

- calculus -
**alia**, Sunday, December 16, 2012 at 11:19pmjust plug pi/4

right

- calculus -
**alia**, Sunday, December 16, 2012 at 11:30pmthnks Reiny.

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