Find an eqn for the tangent line to the graph of f at the point P(pi/4, f(pi/4)) when f(x)= 8 tan x?

Question

A) y= 14x+ 4( 1- pi/2)
B) y= 16x +8(1- pi/2)
C) y=18x +14(1- pi/2)
D) y= 21x +10(1- pi/4)
E) y= 22x+5(1-pi/4)

Answer 1

B) y = 16x + 8(1 - π/2)

f(x) = 8tanx
f'(x) = 8sec²x
f'(π/4) = 8(√2)² = 8*2 = 16

f(x) = 8tanx
f(π/4) = 8tan(π/4) = 8(1) = 8

The equation of the tangent line at (π/4,8) is therefore

y - 8 = 16(x - π/4) = 16x - 4π
y = 16x + 8 - 4π
y = 16x + 8(1 - π/2)

Answer 2

f´(x) = 8*(1+tan^2x) f´(pi/4)=16
f(pi/4)=8

y-8= 16(x-pi/4) y= 16x +8(1-pi/2) B)

Answer 3

f(x) = 8 tan x
f(π/4) = 8 tan π/4 = 8

f `(x) = 8 sec² x = 8 / cos² x
f `(π/4) = 8 / (1/2) = 16

Tangent passes thro` (π/4, 8):-

y - 8 = 16(x - π/4)
y = 16x - 4π + 8
y = 16x + 8(1 - π/2)-----------ANSWER B