Question

how do I find the second derivative on ln(2+sinx)?

Answer 1

\[f(x)=\ln(2+\sin(x)) \\ f'(x)=(2+\sin(x))' \cdot \frac{1}{2+\sin(x)} \text{ by chain rule }\]
find derivative of ln(y) you do derivative of y/y

Answer 2

\[\frac{d}{dx}\ln(y)=\frac{\frac{d}{dx}y}{y}\]

Answer 3

i got cos/2+sin as the first derivative

Answer 4

I think it is right!
\[\frac{ \cos x }{ 1+2 sinx }\]

Answer 5

as first derivative, of course, now you have to apply the rule of derivative of a quotient of two function , please!

Answer 6

namely:
\[\left( \frac{ f }{ g } \right)'=\frac{ f'*g-f*g' }{ g ^{2} }\]

Answer 7

where f=cos x, anf g=1+2 sin x

Answer 8

@jbauer33 try please!

Answer 9

i got -2 sinx - sin^2x - cos^2x/(2+sinx)^2 but i am having trouble simplifying

Answer 10

@Michele_Laino ?

Answer 11

\[\frac{-\sin(x)-2\sin^2(x)-\cos(x)(2\cos(x))}{(2+\sin(x))^2}\]
I think you are missing a 2 in fron of the cos^2 on top

Answer 12

guess what -2sin^2(x)-2cos^2(x)=?
:)

Answer 13

we know sin^2(x)+cos^2(x)=1
so -2sin^2(x)-2sin^2(x)=?

Answer 14

where did the 2 infront of the second cos come from?