How do you find the first derivative of $y = {(\frac{sin x}{1 + cos x})}^{2}$ $y=(sinx/(1+cosx))^2$ ?

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Answer 1 · 2015-05-22T16:33:53

Let's use the chain rule, by naming $u = \frac{sin x}{1 + cos x}$ $u=(sinx)/(1+cosx)$ .

The chain rule states that

$\frac{d y}{d x} = \frac{d y}{d u} \frac{d u}{d x}$ $(dy)/(dx)=(dy)/(du)(du)/(dx)$

Thus,

$\frac{d y}{d u} = 2 u$ $(dy)/(du)=2u$

$\frac{d u}{d x}$ $(du)/(dx)$ - here, we have to apply quocient rule, which states that

Be $y = \frac{f (x)}{g (x)}$ $y=(f(x))/(g(x))$ , $(dy)/(dx)=(f'(x)g(x)-f(x)g'(x))/(f(x))^2$

$(du)/(dx)=(((cosx)(1+cosx))-(sinx)(-sinx))/(1+cosx)^2$

$(du)/(dx)=((cos^2x+cosx)+sin^2x)/(1+cosx)^2$

$(du)/(dx)=(color(green)(cos^2x+sin^2x)+cosx)/(1+cosx)^2$

$(du)/(dx)=cancel(1+cosx)/(1+cosx)^cancel2$

$(du)/(dx)=1/(1+cosx)$

Now, combining $(du)/(dx)$ and $(dy)/(du)$ :

$(dy)/(dx)=(2u)(1/(1+cosx))$

$(dy)/(dx)=2(sinx/(1+cosx))(1/(1+cosx))$

$(dy)/(dx)=color(green)((2sinx)/(1+cosx)^2)$

How do you find the first derivative of y=(sinx/(1+cosx))^2y=(sinx1+cosx)2y=(sinx/(1+cosx))^2?