Question

can someone help me find the derivative of this problem?
f(x)= ln (sqrt of 1+sin^2x)

Answer 1

chain rule practice !!!

Answer 2

Let me help you write it in neat:
\(ln\sqrt{1+sin^2x}\)

Answer 3

what is the derivative of ln(u) ?

Answer 4

1/u

Answer 5

almost: u is a function, so the chain rule applies:

implicitly we have: ln(u) derives to u'/u

but u = sqrt(v), therefore what is u'?

Answer 6

edit:

u' = v' / 2sqrt(v)

but v = 1 + sin^2(x), so what is v'?

ln(u(v(x))) = 1/u * 1/2sqrt(v) * v'

or: ln(u(v(x))) = v'/ 2usqrt(v)

Answer 7

uhm well v'= 1/2(1+sin^2x)^-1/2 (2)(sinx)(cosx)

Answer 8

its a chain rule practice problem ... just gonna be messy messy messy

Answer 9

\[\frac{d}{dx}ln(\sqrt{1+sin^2x})\]
\[\frac{\frac d{dx}(\sqrt{1+sin^2x})}{\sqrt{1+sin^2x}}\]
\[\frac{\frac d{dx}(1+sin^2x)}{2\sqrt{1+sin^2x}\sqrt{1+sin^2x}}\]
\[\frac{\frac d{dx}(1)+\frac d{dx}(sin^2x)}{2(1+sin^2x)}\]
\[\frac{2sin(x)\frac d{dx}(sinx)}{2(1+sin^2x)}\]
\[\frac{2sin(x)cos(x)}{2(1+sin^2x)}\]

Answer 10

spose we can cancel the 2s and such but i think thats about it

Answer 11

okay wait but why divide by sqrt of 1+ sin^2x

Answer 12

no actually how did you get sqrt of 1+sin^2x in the numerator?