Question

Find the derivative of cosx/(1+sinx)

Answer 1

The answer is
\[\frac{ 1 }{1 -sinx }\]

Answer 2

The answer I got was
\[\frac{ -1 }{ 1+sinx}\]

Answer 3

My question is are those the same thing?

Answer 4

They are not.
\[\frac{1}{1-\sin x}=\frac{-1}{-(1-\sin x)}=\frac{-1}{-1+\sin x}\neq\frac{-1}{1+\sin x}\]

Answer 5

Oh ok

Answer 6

So its a mistake to factor out a -1 in the denominator in this situation?
\[\frac{ -sinx -1}{ (sinx+1)^2 }\]

Answer 7

i mean numerator*

Answer 8

Your answer is correct.\[\begin{align*}
\frac{\mathrm d}{\mathrm dx}\left[\frac{\cos x}{1+\sin x}\right]&=\frac{\dfrac{\mathrm d}{\mathrm dx}[\cos x](1+\sin x)-\dfrac{\mathrm d}{\mathrm dx}[1+\sin x]\cos x}{(1+\sin x)^2}\\[1ex]
&=\frac{-\sin x(1+\sin x)-\cos^2x}{(1+\sin x)^2}\\[1ex]
&=\frac{-\sin x-(\sin^2x+\cos^2x)}{(1+\sin x)^2}\\[1ex]
&=-\frac{\sin x+1}{(1+\sin x)^2}\\[1ex]
&=-\frac{1}{1+\sin x}
\end{align*}\]

Answer 9

Should I type out the work I have if it's confusing to understand what i'm asking?

Answer 10

Oh really? The derivative calculator had a different answer and I was confused as to why.

Answer 11

Yup. You can always integrate to check.
\[\begin{align*}
-\int\frac{\mathrm dx}{1+\sin x}&=-\int\frac{1-\sin x}{\cos^2x}\,\mathrm dx\\[1ex]
&=\int\left(\sec x\tan x-\sec^2x\right)\,\mathrm dx\\[1ex]
&=\sec x-\tan x+C\\[1ex]
&=\frac{1-\sin x}{\cos x}+C\\[1ex]
&=\frac{1-\sin^2x}{\cos x(1+\sin x)}+C\\[1ex]
&=\frac{\cos^2x}{\cos x(1+\sin x)}+C\\[1ex]
&=\frac{\cos x}{1+\sin x}+C
\end{align*}\]

Answer 12

\(\dfrac{d}{dx}\dfrac{\cos x}{1 + \sin x} =\)

\(=\dfrac{d}{dx}[\cos x(1 + \sin x)^{-1}]\)

\(=\cos x \dfrac{d}{dx}(1 + \sin x)^{-1} + (1 + \sin x)^{-1} \dfrac{d}{dx} \cos x\)

\(= \cos x [-(1 + \sin x)^{-2}]\cos x + (1 + \sin x)^{-1}(- \sin x)\)

\(= -\dfrac{\cos ^2 x}{(1 + \sin x)^2} - \dfrac{\sin x}{1 + \sin x} \)

\(= -\dfrac{\cos ^2 x}{(1 + \sin x)^2} - \dfrac{\sin x(1 + \sin x)}{(1 + \sin x)^2} \)

\(= \dfrac{-\cos ^2 x - \sin x - \sin^2 x}{(1 + \sin x)^2}\)

\(= -\dfrac{\sin^2 x + \cos ^2 x + \sin x}{(1 + \sin x)^2}\)

\(= -\dfrac{1 + \sin x}{(1 + \sin x)^2}\)

\(= -\dfrac{1 }{1 + \sin x}\)

Answer 13

Alright thanks guys :D

Answer 14

You're welcome.