# derivative of sin x°

• Mar 14th 2009, 06:53 AM
ycdfd
derivative of sin x°
Hi,
I just want to know if there is a difference between the derivative of sin x and sin x°, where x is in radians & x° is in degree.
My textbook says the derivative of sin x is cos x, if and only if x is in RADIANS. But now I am asked to work out the derivative of sin x°. I think d(sin x°)/dx shouldn’t equal to cos x° because x° is not in radians. If I am making a right point here, what is the derivative of sin x°?
Thank you!
• Mar 14th 2009, 07:06 AM
atac777
Convert x degrees to radians and take the derivative of that?

y' = $\frac{d}{dx}\sin\frac{x\pi}{180}$.

Degrees and radians are interchangeable, so you can substitute one for another in a trig equation, so long as they come out to be the same when converted back.

Product rule inside a quotient rule inside a chain rule.

This is a complete shot in the dark, so I may be wrong on this.
• Mar 14th 2009, 07:22 AM
atac777
Got y' = 0 as my answer. Didn't even need the chain rule. When I got to that point, I just had $y' = \sin(0)$.
• Mar 14th 2009, 07:27 AM
metlx
f'(x) = (π/180)cos(πx/180)

i think..(not so confident anymore after having several posts edited by the mods :P)
• Mar 14th 2009, 07:36 AM
ycdfd
Quote:

Originally Posted by metlx
f'(x) = (π/180)cos(πx/180)

i think..(not so confident anymore after having several posts edited by the mods :P)

i got the same answer too
thank you guys for all the help
• Mar 15th 2009, 03:39 PM
atac777
Ignore this post. I was an idiot and mixed up the quotient rule and product rule signs.