derivative of sin x°
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°?
Really need your help
Convert x degrees to radians and take the derivative of that?
y' = .
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.
Got y' = 0 as my answer. Didn't even need the chain rule. When I got to that point, I just had .
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
Originally Posted by metlx
thank you guys for all the help
Ignore this post. I was an idiot and mixed up the quotient rule and product rule signs.