derivatives...?

Question

y=cosx/(1+sinx)

x+siny=xy

y=(sinx/1+cosx)^2

Answer 1

y'={-sinx(1+sinx)-cos^2 x}/(1+sinx)^2



1+y'cosy=1y+y'x


y'=2(sinx/1+cosx) .{ (cosx (1+cosx)(-sin^x)/(1+cosx)^2}

Answer 2

You'll get answers to the first and third, cuz they look pretty straight forward. But the second is different.

This is a relation equation and not a function. I assume you're using implicit differentiation. Look up those vocab words in your text to see what I mean, and to be able to recognize the differences.

gawd, I'm SUCH a math teacher. Ok, here goes.

To find y ' (which is the derivative of y with respect to x) you take the derivative one term at a time, remembering to chain whenever you take a derivative of y. Let's do these one term at a time:

the derivative of x with respect to x is 1
the derivative of sin (y) with respect to x is cos (y) TIMES the derivative of y with respect to x. We write, y ' cos (y)
the derivative of xy with respect to x requires the product rule.
(xy) ' = xy ' + yx '

but x ' is 1, we already decided that. so,
(xy) ' = xy ' + y

putting all that together, we have
1 + y ' cos (y) = xy ' + y

Use algebra to isolate the y ' by iteself, and you're done.

y ' = (y - 1) / (cosy - x)

Answer 3

1] apply the division rule

dy/dx = [ -sinx(1+sinx) - cosx(cosx)] / (1+sinx)^2

-1/ [1+sinx]

2] diffrentiate through out w.r.t x
you get ,
1+cosy * dy/dx = y+ x*dy/dx.
then get the value of dy/dx.
3]
similar to q1]