find deriv of following

G

Guest

Guest
Hi Guys,
I may have posted a problem like this before but I searched and can't find it.
Problem is.

Find deriv of y = f(x) = [cos(x)-cot(x)]/csc(x)

ok I know I can simplify this some using identities from trig,

so wouldnt cot(x)/csc(x) = cos(x)?
and cos(x)/csc(x) = sin(x)cos(x)?

Now, how would I take these values and diff them?

I have in my notes where I tried this and came up with -2(cos(x))^2 - 1.
Although I can't find my work for it, so I don't remember how I came up with that. Can you guys give me a shove towards solving this? Thanks again.
Chris
 
[cos(x)-cot(x)]/csc(x) = sin(x)cos(x)-cos(x)

Now use the product rule and it'll fall into place.
 
Thanks Galactacus,

How does this look?

d/dx cos(x)sin(x) - d/dx cos(x),

= d/dx cos(x)sin(x) + sin(x)

so using prod rule u = cos(x), v = sin(x)

=cos(x) d/dx sin(x) + sin(x) d/dx cos(x) + sin(x),

= cos(x)^2 + sin(x) d/dx cos(x) + sin(x),

= cos(x)^2 - sin(x)^2 + sin(x)?
 
y=[cosx-cotx]/cscx but cotx= cosx/sinx and cscx=1/sinx substitute
y= cosxsinx - cosx
y'= cosx^2-sinx^2x+sinx
y'=cos^2x-[1-cos^2x]+sinx
y'=2cos^2x+[sinx-1]
Arthur
 
Arthur,

Im with you til the third line.... Can you explain to me how you got the following?

y'=cos^2x-[1-cos^2x]+sinx
y'=2cos^2x+[sinx-1]

Thanks for the reply...

Chris
 
y= cos(x)sin(x) - cos(x)

d(cos(x)sin(x)) =
cos(x)d(sin(x))+sin(x)d(cos(x))=
cos(x)²-sin(x)²

sin(x)²=1-cos(x)²
cos(x)²-sin(x)²=
cos(x)²-(1-cos(x)²) =
2cos(x)²-1

d(-cos(x)) =
sin(x)

put the parts together for
y'=2cos(x)²+[sin(x)-1]
 
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