derivative of trigonemtric function?

Question

y=(sinx/1-cosx)^2 derivative of this function

Answer 1

You can of course use a brute force approach to get the derivative but a little finesse often can simplify things. You have:

y=(sin(x)/1-cos(x))^2 = (sin(x)^2)/(1-cos(x))^2

Since sin(x)^2 = 1 - cos(x)^2 it follows that:

y = (1 - cos(x)^2)/(1 - cos(x))^2

By doing a little algebra up fornt, the final result is quite a bit simpler than some of the others given.
Factoring the numerator:

y = (1 - cos(x))*(1 + cos(x))/(1- cos(x))^2 = (1 + cos(x))/(1 - cos(x))

Now your function is of the form

y = (1 + u)/(1 - u) where u = cos(x)

dy/dx = dy/du * du/dx = (1/(1 - u) + (1 + u)/(1 - u)^2)(-sin(x))

dy/dx = -2*sin(x)/(1 - u)^2 = -2*sin(x)/(1 - cos(x))^2

By doing a little algebra up front, the final result is quite a bit sinpler than those above.

Answer 2

= sinx^2(1-cosx)^-2
=2sinx(1-cosx)^-2 -2sinx^2(1-cosx)^(-3)(sinx)

I wouldn't leave it in that form...you should simplify

And btw sinx^2 + cosx^2 = 1 NOT what that other person said

Answer 3

Sorry, (1-cosx)^2 does not equal (sinx)^2

y=(sinx/(1-cosx))^2
y=(sinx)^2 / (1-cosx)^2
y = (sinx)^2 / (1 - 2cosx + (cosx)^2)

Use quotient rule for derivatives:

y' = [(1 - 2cosx + (cosx)^2)(2sinxcosx) - ((sinx)^2)(2sinx -2sinxcosx)) ] / [ (1-cosx)^4]

Answer 4

sinx/(1-cos x) = sin x (1+ cos x)/(1- cos^2 x)
= (1+cos x)/ sin x
= 2 cos^2 x/2/ 2 sin x/2 cos x/2
= 2 cot x/2
so y = (2 cot x/2)^2
now dy/dx = -4 cot x/2 cosec^2 x/2 d/dx(x/2) = -2 cot x/2 cosec^2 x/2

using chain rule d/dx(f = df/dt.dt/dx where t = cot x/2

Answer 5

y=(sinx/1-cosx)^2
{[2(sinx/2)(cosx/2)]/[2sin^2(x/2)]}^2
[(cosx/2)/(sinx/2)]^2
(cotx/2)^2
diff wrt x
2cotx/2*(-cosec^2(x/2))*1/2
-cotx/2*cosec^2(x/2)*

Answer 6

2(sinx/(1-cosx)[(1-cosx)*cosx)
-sinx(sinx)/(1-cosx)^2]
=2sinx[(1-cosx)cosx+sin^2x)
/(1-cosx)^3