Derivatives

[ohai]

Please, can anyone explain to me how to do the following problems? Thanks.

Directions: find dy/dx
1. y= ((sinx)/(1+cosx))squared

2. y= (cscx + cotx)to the negative one power

3. The position of a particle miving laong a cordinate line is s=the square root of the quantity 1+4t, with s in meters and t in seconds. Find the particle's velocity and acceleration at t=6secs.

4. Find the equations of the lines that are tangent and normal to the curve y=(the square root of two)cosx at the point (pi/4, 1).

Thanks again,
J

 

[Guest]

find dy/dx
1. y= ((sinx)/(1+cosx))squared

y= p^2 where p = u/v

p = (u/v) where u = sin x and v = 1+cos x

du/dx = cos x

dv/dx = -sin x

dy/dp = 2p


dp/dx = (vdu/dx - u dv/dx) / v^2

now have a go................

 

[Guest]

dp/dx = (vdu/dx - u dv/dx) / v^2

dp/dx = ( (1+cos x) (cos x) - (sin x) ( -sin x)) / ((1+cos x)^2)

= (( cos x + (cos x) ^2) +(sin x)^2)/ ((1+cos x)^2)
= cos x +1 /(1+cos x) (1+cos x)

= 1 / (1+cos x)

= (1+cos x) ^-1

now dy/dx = dp/dx . dy/dp

= (1+cos x) ^-1 . 2p where p = (u/v) where
u = sin x and v = 1+cos x

dy/dx = (1+cos x) ^-1 . 2 (sinx/(1+cos x)

= 2 sin x (1+cos x)^-2

does this help?