# Thread: derivative of sin x°

1. ## 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!

2. 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.

3. Got y' = 0 as my answer. Didn't even need the chain rule. When I got to that point, I just had $y' = \sin(0)$.

4. f'(x) = (π/180)cos(πx/180)

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

5. 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

6. Ignore this post. I was an idiot and mixed up the quotient rule and product rule signs.