The derivative of \(e^x \cos x\) with respect to cosx is : a)cosx - sinx b)cosx+sinx c)sinx-cosx d)\(e^x\)(sinx+cosx)
i guess i do not understand what it means when you say "with respect to cosine"
should be option d.
@satellite73 it means instead of finding dy/dx....we have to find dy/d(cosx).
It says, its option a P.S. I guess sth is wrong with the question! But may be i am wrong!
but e^x should be in the answer!
And that makes me think that something is wrong with qsn or answer!
i think so too....and the answer i got is e^x (1 - tanx).
Why is wolf not helping? http://www.wolframalpha.com/input/?i=differentiate+e%5Ex+cosx+with+respect+to+cosx
i guess it only does standard differentiation....ie, w.r.t x
oh sorry...i wrote the wrong answer before. it should be e^x (1 - cotx)
if it is really wrt cosine, then it should be \(e^x\)
@satellite73 how come?
because it says "with respect to cosine" unless you want to write cosines in terms of \(e^x\) using complex numbers i guess i should be quiet because i am not really sure what "with respect to cosine" means in this case
say: cos(x) = u x = arccos(u) e^(arccos(u)) u
maybe amistre knows
dy/d(cosx) can also be written as dy/dx the whole upon d cosx / dx.......now it is easier to solve.
ooh is that how it works?
i dunno if my method is right, but its just a hunch ... prolly wrong tho :)
I get this:\[-\frac{e^x(\cos x + \sin x)}{\cos^2 x \sin x}\]
Guys! This shouldn't be much complex! It doesn't fall in that category in my text book!
@satellite73 no!! @ujjwal i know! and i'm telling u....i'm pretty confident about my answer.
still lost i will shut up
my idea gets me e^x(1-cotx)
let y = e^x * cos x so dy/dx = e^x cosx - e^x sinx = e^x ( cosx - sinx).........(i) now, d cosx/ dx = -sinx.............(ii) now do (i) by (ii) to get the final answer.
You missed dividing by (e^x)^2 in eqn 1 @Vaidehi09
why do we have to do that? the product rule says: (uv)' = uv' + vu'
Oh! yes.. I forget small things!
so the answer is undoubtedly e^x ( 1 - cotx).
Yes! the answer is none of the given options! LOL...
lol..yea!
Question closed! thanks everyone!
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