Question

Can anyone help me with this?
Find the derivative:

f(x)=ln(sin(e^x))+(1/2(tan))[(ln(x+1))]

Answer 1

It's going to be a pain in the retrice but sure.

Break that pellet into two derivatives. Let's do this pellet.

1) ln(sin(e^x)) derived is [ 1 / ( sin(e^x)) ] * [ sin(e^x) ] ' = [ 1 / ( sin(e^x)) ] * [ cos(e^x) *e^x] = ctg (e^x) * e^x

2) (1/2(tan))[(ln(x+1)) derived is ...

wait, did you mean 1/2 * tan ( ln(x+1) ) by any chance ?

Answer 2

1/2tan[ln(x+1)]

Answer 3

\[f(x)=\ln(\sin(e^{x})) \]
let u=sin(e^x) to get f(x)=lnu
\[f'(x)= \frac{ f'(u) }{ f(u) }\]
\[f'(x)=\frac{ e^{x}\cos(e^{x}) }{ \sin(e^{x}) }\]

\[\frac{ cosx }{ sinx }=cotx\]

\[f'(x)=e^{x}\cot(e^{x})\]

now another function
\[g(x)=\frac{ 1 }{ 2 }\tan(\ln(x+1))\]
\[g'(x)=\frac{ \sec^{2}\ln(x+1) }{ 2(x+1)}\]

add them together to get the differential

\[f'(x)=e^{x}\cot(e^{x}) + \frac{ \sec^{x}\ln(x+1) }{ 2(x+1) }\]

Answer 4

to differentiate the tan function I let ln(x+1)=u
so I have
\[g(x)=\frac{ 1 }{ 2 }\tan(u)\]
\[g'(x)=\frac{ 1 }{ 2 }\sec^{2}u \]

then I multiply by the differential of u

Answer 5

oops that should be u in the brackets
g(u)
g'(u)

Answer 6

hope that helps

Answer 7

wait so what's the answer

Answer 8

i think i got something different

Answer 9

hope that helps

Answer 10

I gave you the answer

Answer 11

http://prntscr.com/6hb0bp

Answer 12

oops didn't see that my chat is lagging for some reason

Answer 13

need me to explain anything?

Answer 14

Yes I am having trouble with the trapezoid and simpsons rule i was wondering if you could help me on that

Answer 15

I meant on this question lol
post your next question in a new question and I will try to help

Answer 16

Ok I will post a new question.