what's the derivative of y=ln square root of 1+ sin2x

Question

anonymous · Answer

$$y= \ln \sqrt{1+\sin2x}$$

amistre64 · Answer

logs pull out exponents ...

ln(a^b) = b ln(a)  that should at least simplify the process

amistre64 · Answer

take the "e" of both sides, that will undo the ln

amistre64 · Answer

since the constant is trivial ....
$$y= ln(x)$$
$$e^y= e^{ln(x)}$$
$$e^y= x$$
$$y'e^y= x'$$
$$y'= e^{-y}$$

amistre64 · Answer

i feel like if made an error ....

amistre64 · Answer

in this case, i was trying to signify x' as the derivative of the innards  ... so x' not equal to 1.  bad choice of variable name on my part :)

anonymous · Answer

why you took 'e'? no need to cancel out the ln .. can't you just do this y=ln (1+sin2x)^1/2

amistre64 · Answer

yeah, i was thinking integration with that .... i blame these tired old eyes :/

amistre64 · Answer

good catch tho ...
$$y= \ln \sqrt{1+\sin2x}$$
$$y= \frac12\ln (1+\sin2x)$$

anonymous · Answer

didn't get what you did..

amistre64 · Answer

i just made it less scary by pulling out the sqrt of it.

anonymous · Answer

how it came 1/2 ln ...

amistre64 · Answer

$$\sqrt{n}=n^{1/2}$$
$$log(a^b)=b~log(a)$$

amistre64 · Answer

these are algebraic propeties

amistre64 · Answer

and the derivaitive of a log is:$$D[log(u)]=\frac{u'}{u}$$

anonymous · Answer

ohhhh this point:) thaanks:D

anonymous · Answer

you made it easy sir

amistre64 · Answer

:)  i do get there in the end

anonymous · Answer

hats off to you