Question

Result of the derivative of `-(cos x)/(sin x)^2` ?

I tried to solved it but i'm not sure if it's wright. Would really appreciatte if someone could give the correct result

Answer 1

`d/dx ( -cosx/sin^2x) = (1+cos^2x)/sin^3x`

Explanation:

Using the quotient rule:

`d/dx ((f(x))/(g(x))) = (g(x) (df)/dx - f(x) (dg)/(dx))/(g^2(x))`

we have:

`d/dx ( -cosx/sin^2x) = -( sin^2x d/dx cosx - cosx d/dx sin^2x)/sin^4x`

`d/dx ( -cosx/sin^2x) = ( sin^3x + 2sinx cos^2x )/sin^4x`

simplify:

`d/dx ( -cosx/sin^2x) = ( sin^2x + 2 cos^2x )/sin^3x`

We can write the results in different ways:

`d/dx ( -cosx/sin^2x) = 1/sinx + 2cos^2x /sin^3x = cscx+2secx cot^3x`

or:

`d/dx ( -cosx/sin^2x) = ( sin^2x +cos^2 x+ cos^2x )/sin^3x = (1+cos^2x)/sin^3x`

`d/dx ( -cosx/sin^2x) = (1+cos^2x)/(sinx(1-cos^2x))`

`d/dx ( -cosx/sin^2x) = 1/sinx(1+(1+cos2x)/2)/(1-(1 + cos2x)/2)`

`d/dx ( -cosx/sin^2x) = 1/sinx(3+cos2x)/(1- cos2x)`

Answer 2

`y=color(red)('-')`, then,
`color(blue)((1+cos^2x)/sin^3x)`

Explanation:

`(1)y=color(red)(-)cosx/(sinx)^2=-1/sinx*cosx/sinx=-cscx*cotx`
`(dy)/(dx)=-[cscx*d/(dx)(cotx)+cotx*d/(dx)(cscx)]`
`(dy)/(dx)=-[cscx*(-csc^2x)+cotx*(-cscxcotx)]`
`(dy)/(dx)=color(red)(csc^3x+cscx*cot^2x=cscx(csc^2x+cot^2x))`
Second method:
`y=-[(cosx)/sin^2x]`
`(dy)/(dx)=-[((sin^2x)d/(dx)(cosx)-cosxd/(dx)(sin^2x))/(sin^2x)^2]`
`=-[(sin^2x(-sinx)-cosx*2sinxcosx)/(sin^4x)]`
`=[(sin^3x+2sinxcos^2x)/(sin^4x)]=(sin^2x+2cos^2x)/sin^3x=color(blue)((1+cos^2x)/sin^3x)`