using the bernoulli's method is much faster .
1 . Take the Ln from two sides.
2 . Now : Derivinting from two sides :
get
Ln(y) = Ln( ) =>
Ln(y) = [Ln(x) +Ln( = >
= ( + =>
y ( + =>
( + =>
set = >
( +
I have this question :
Find out derivative : y = √(x √(x + sineⁿ))
{ I use : n = x }
I named : u = x + sineⁿ
so,
y' = ½ (x + √y)ª · (1 + ½ · uª)· u'
{ I use : a = - ½ }
And : u' = 1 + (sineⁿ)'
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...