chain rule

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Truscuit

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find dy/dx of y=(sinx/1+cosx)^2

so far got this y=u^2
y'=2u which eguals 2(sinx/1+cosx)

u=(sinx/1+cosx)
u'= [(1+cosx)(cosx)-(sinx)(-sinx)]/(1+cosx)^2

dy/dx= (2sinx/1+cosx)(cosx+cos^2x+sin^2x/(1+cosx)^2

i was wondering if i did it so far right and if so can it be simplified any further

thanks
 
Note that \(\displaystyle \frac{sin(x)}{1+cos(x)}=tan(\frac{x}{2})\)

So, find the derivative of \(\displaystyle tan^{2}(\frac{x}{2})\)

That may be a little easier.
 
Truscuit said:
find dy/dx of y=(sinx/1+cosx)^2

so far got this y=u^2
y'=2u which eguals 2(sinx/1+cosx)

u=(sinx/1+cosx)
u'= [(1+cosx)(cosx)-(sinx)(-sinx)]/(1+cosx)^2

dy/dx= (2sinx/1+cosx)(cosx+cos^2x+sin^2x/(1+cosx)^2

i was wondering if i did it so far right and if so can it be simplified any further

thanks
Click to expand...

\(\displaystyle \frac{d}{dx}\ \left [ \frac{sin(x)}{1+ cos(x)}\right ]^2\)

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

\(\displaystyle = \ 2\left [ \frac{sin(x)}{1+ cos(x)}\right ]\frac{1+cos(x)}{[1+cos(x)]^2}\)

\(\displaystyle = \ 2\left [ \frac{sin(x)}{[1+ cos(x)]^2}\right ]}\)
 
Subhotosh Khan
can you explain in more on how you came to the anwer
like what happened to the (-sin)(-sin) and how you combined the to ge the answer
\(\displaystyle \frac{d}{dx}\ \left [ \frac{sin(x)}{1+ cos(x)}\right ]^2\)

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

\(\displaystyle = \ 2\left [ \frac{sin(x)}{1+ cos(x)}\right ]\frac{1+cos(x)}{[1+cos(x)]^2}\)

\(\displaystyle = \ 2\left [ \frac{sin(x)}{[1+ cos(x)]^2}\right ]}\)
 
The derivative of \(\displaystyle tan^{2}(\frac{x}{2})\)

Chain rule is basically the derivative of the inside times the derivative of the outside...to put it simply.

The derivative of x/2 is 1/2.

The derivative of outside is \(\displaystyle 2tan(\frac{x}{2})\)

The derivative of the inside is the derivative of tan(x), which is \(\displaystyle sec^{2}(\frac{x}{2})\)

Put it all together and get \(\displaystyle \frac{1}{2}\cdot 2\cdot tan(\frac{x}{2})sec^{2}(\frac{x}{2})=tan(\frac{x}{2})sec^{2}(\frac{x}{2})=\frac{sin(\frac{x}{2})}{cos^{3}(\frac{x}{2})}\)

This is an identity equal to \(\displaystyle \frac{2sin(x)}{(1+cos(x))^{2}}\)

Either way would be OK. But you may want to best go with SK's method.
 
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