Differention?

Question

For the curve, find the equation of the tangent to the curve.
y = 2cos4x at x=pi/8 (pi is 3.14159...)

I need two answers involving pi and the correct one
May use product rule.

Answer 1

Differentiating, you get y' = -8 sin 4x. No product rule required.

At x=pi/8, you have y' = -8 sin(4pi/8) = -8 sin(pi/2) = -8.

Using the point-slope form,

y-y0 = m(x-x0)

with x0=pi/8, y0=2 cos(4pi/8)=2 cos(pi/2) = 0, and m=-8, you get

y=0 - 8(x-pi/8) = -8x + pi.

Answer 2

y = 2cos4x remember if y= cosx, dy/dx is -sinx.

if y = cos U, dy/dx = -sinU dU/dx, where

U is a function of 'x'. For example, U = f(x) = 4x or 4/x or 5x^3 and

so on.

if y = n cosU , dy/dx = -n sin U dU/dx, where 'n' is number such

as 2 or 2/3 or 6 or 7 and such.

so if y = 2cos 4x, dy/dx = -2 sin4x d/dx 4x

= -2 sin4x 4

= -8 sin 4x

when x = pi/8,

dy/dx = -8 sin 4 pi/8

= -8 sin pi/2

notice the use of 'pi' in maths. when we calculate the area of

circle, 'pi' is 22/7 or 3.14.

in this case, assuming 'pi' as radian or 180* is more appropriate.

so pi/2 = 180*/2 = 90*

sin90* is 1.

so dy/dx = -8

when x = pi/8 , y = 2 cos 4 pi/8

= 2 cos pi/2

cos pi/2 is cos 90*. cos 90* is equal to zero.

so y= 0.

( pi/ 8 , 0 ) is the point of contact to the curve .

equation of tangent line is y - y1 = m ( x- x1 ), where 'm' is dy/dx.

y - 0 = -8 ( x - pi/8 )

y = -8x + pi

y = -8x + pi is equation of the tangent line to the curve at x = pi/8

Answer 3

y = 2 cos (4x)
y' = -8 sin (4x)
so, gradient of curve at x = pi/8 is -8 sin (pi/2) = -8.
and it goes through the point (pi/8 , 0)

So the equation is:

y = -8 (x - pi/8)

so y = pi - 8x


You don't need the product rule, you need the chain rule.
You should try www.quickmath.com if you get stuck on calculus.

Answer 4

y = 2cos4x
y' = 2(-sin4x)(4)
y' = -8sin4x

When x = pi/8, y = 0 and y' = -8.

So the equation of the tangent at x = pi/8 is:
y = mx + c
0 = (-8)(pi/8) + c
c = pi

y = -8x + pi => 8x + y = pi