1. Derivatives and logarithm

I have this question :

Find out derivative : y = √(x √(x + sineⁿ))

{ I use : n = x }

I named : u = x + sine

so,

y' = ½ (x + √y· (1 + ½ · )· u'

{ I use : a = - ½ }

And : u' = 1 + (sine)'

But I have many problems with the last number. I Think it would be something like this :

Ln y = x Ln e
1/y· y' = x/sine· sine' + Ln sine
y'/y = (x cose + sine · Ln sine) / sine
y' = sineⁿ‾¹ (x cose + sine · Ln sine)

Finally I think there is something wrong, cause at last, I find out a very complex expression...

2. using the bernoulli's method is much faster .

1 . Take the Ln from two sides.
2 . Now : Derivinting from two sides :

get $\displaystyle e ^ n = \theta$

Ln(y) = $\displaystyle \frac{1}{2}$ Ln($\displaystyle x\sqrt{x+sin\theta}$) =>

Ln(y) = $\displaystyle \frac{1}{2}$[Ln(x) +Ln($\displaystyle \sqrt{x+sin\theta})]$ = >

$\displaystyle \frac{y'}{y}$ = $\displaystyle \frac{1}{2}$($\displaystyle \frac{1}{x}$ + $\displaystyle \frac{1+cos\theta}{2\sqrt{x+sin\theta}})$ =>

$\displaystyle y' =$ y $\displaystyle \frac{1}{2}$($\displaystyle \frac{1}{x}$ + $\displaystyle \frac{1+cos\theta}{2\sqrt{x+sin\theta}})$ =>

$\displaystyle y' =$ $\displaystyle \sqrt{x\sqrt{x+sin\theta}}$($\displaystyle \frac{1}{x}$ + $\displaystyle \frac{1+cos\theta}{2\sqrt{x+sin\theta}})$$\displaystyle \frac{1}{2} => set \displaystyle \theta = e ^ n = > \displaystyle y' = \displaystyle \sqrt{x\sqrt{x+sin(e ^ n)}}(\displaystyle \frac{1}{x} + \displaystyle \frac{1+cos(e ^ n)}{2\sqrt{x+sin(e ^ n)}})$$\displaystyle \frac{1}{2}$