Find the equation of the tangent line to the curve y=4tanx at the point (pie/4,4) . The equation of this?

Question

tangent line can be written in the form y=mx+b where m is.... and b is???

Answer 1

f(x) = 4 tanx

f `(x) = 4sec²x = 4(1/cos² x)

f ` (π/4) = 8

Gradient m = 8
Curve passes thro` (π/4,4) where 4 = 4tan(π/4)

Equation of tangent thro` (π/4,4) with m = 8 is given by:-

y - 4 = 8(x - π/4)

y = 8x - 6.28 + 4

y = 8x - 2.28

Answer 2

what's the curve equation to discover the tangent line at a level (25,5) on it? EDIT: [As in line with the added information, which i ought to relook purely after about 9 hours of previously presentation] a million) Differentiating the given one, dy/dx = a million/(2?x) 2) At x = 25, dy/dx = a million/(2?25) = a million/10; that's the slope of the tangent line on the given element; that's by technique of the geometrical definition of differentiation] 3) using Slope-element type, the equation of the tangent line at (25,5) is: y - 5 = (a million/10)(x - 25) increasing and simplifying, the equation is: x - 10y + 25 = 0

Answer 3

slope of the curve is given by

dy / dx = 4sec^2 x

slope at the point (pi/4 , 4) = 4 * sec^2 pi/4 = 8

hence the tangent at that point is of the form
(tangent has the same slope as the curve)

y = 8x + c

now this eqn passes thru the point (pi/4 , 4)

4 = 8 * pi/4 + c
c = 4 -2pi


eqn is

y = 8x + 4 - 2pi

Answer 4

y = 4tanx
dy/dx = 4sec²x = 4sec²(π/4) = 4(√2)² = 4*2 = 8

The slope m of the tangent line at the point (π/4,4) is 8.
The equation of the line is:

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