Question

find the derivative
f(x)=cotx/(1+cscx)
iget -csc^2x-csc^3+cscxcot^2x/(1+cscx)^2

but my book says the answer is -cscx/(1+cscx)
how ca get that?

Answer 1

use d quotient rule

Answer 2

f'(x) = -sinx (1+sinx) - (cosx)(cosx) / (1+sinx)^2 = (-sinx - sin^2 x - cos^2 x) / (1+sinx)^2 = (-sinx -1) / (1+sinx)^2 = - (1+sinx) / (1+sinx)^2 = -1 / (1+sinx)

Answer 3

I did useth qtientrule how ca isipliy

Answer 4

@misty1212 what do u think about this?

Answer 5

Hi are you still there?

Answer 6

I think I have figured it out

Answer 7

\[\cot^2(x)*\csc(x)-\csc^2(x)*\csc(x)+1/\csc^2(x)+1\]

Answer 8

\[(\csc(X)).\frac{ \cot^2(x)-\csc^2(x) }{ \csc(x)+1 }= \frac{ \csc(x) *-1 }{ \csc(x)+1 }\]

Answer 9

and dat is your answer

Answer 10

Notice that the answer obtained by the posts above are the same
\[ -\frac{1}{\sin
(x)+1}=-\frac{1}{\frac{1}{\csc
(x)}+1}=-\frac{1}{\frac{\csc
(x)+1}{\csc (x)}}=-\frac{\csc
(x)}{\csc (x)+1}
\]