if the value of x is in degrees what is the derivative of f(x)=sin(x)


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Question Number 90881 by M±th+et+s last updated on 26/Apr/20

$i f t h e v a l u e o f x i s i n d e g r e e s w h a t i s$ $t h e d e r i v a t i v e o f f (x) = s i n (x)$
Commented by mr W last updated on 28/Apr/20

$w e h a v e d i f f e r e n t u n d e r s t a n d i n g :$ $i f x i s i n r a d, t h e s l o p e \frac{d y}{d x} i s i n u n i t$ $[1 / r a d] . b u t i f x i s i n d e g r e e, t h e$ $s l o p e i s i n u n i t [1 / d e g r e e] . b o t h s l o p e s$ $m u s t n o t e q u a l i n v a l u e, b u t e q u a l i n$ $t h e p h y s i c a l s e n s e .$ $w h e n w e d e s c r i b e t h e m o t i o n o f a c a r$ $w i t h y = f (t) = s i n (t) w h e r e t i s i n s e c o n d,$ $t h e s l o p e \frac{d y}{d t} w h i c h i s t h e s p e e d i n u n i t$ $[m / s] . b u t i f t i s i n m i l e, t h e s l o p e \frac{d y}{d t}$ $i s a l s o t h e s p e e d, j u s t i n a n o t h e r$ $u n i t [m / m i l e] . b o t h a r e o f d i f f e r e n t$ $v a l u e s, b u t m e a n t h e s a m e p h y s i c a l$ $s e n s e . t i n f (t) c a n b e i n s e c o n d o r m i l e,$ $x i n s i n (x) c a n b e i n r a d o r d e g r e e,$ $w h i c h {d o e n}^{'} t a f f e c t h o w w e c a l c u l a t e$ $\frac{d y}{d t} o r \frac{d y}{d x} . t h i s i s m y u n d e r s t a n d i n g,$ $m a y b e c o r r e c t o r m a y b e n o t .$
Commented by mr W last updated on 26/Apr/20

$\frac{d y}{d x} = \cos (x) n o m a t t e r i f x i s i n r a d o r$ $i n d e g r e e .$ $t h i s i s t h e s a m e a s i f s = f (t),$ $v = \frac{d s}{d t} = f^{'} (t), n o m a t t e r i f t i s i n h o u r$ $o r s e c o n d .$
Commented by M±th+et+s last updated on 26/Apr/20

$t h a n k y o u s i r f o r e x p l a i n t h a t$
Commented by MJS last updated on 26/Apr/20

$Sir, this obviuosly is wrong$ $the derivate is the slope of the tangent$ $\frac{d}{d x} [\sin x^{rad}] = \cos x^{rad}$ $but the slope of \sin x ° is not as steep$ $\frac{d}{d x} [\sin x °] = \frac{π}{180} \sin x °$ $generally if f (x °) is any trigonometric$ $function$ $\frac{d}{d x} [f (x °)] = \frac{π}{180} f^{'} (x °)$
Commented by mr W last updated on 26/Apr/20

$i f x i n f (x) i s i n d e g r e e, t h e n t h e x i n$ $d x i s a l s o i n d e g r e e .$ $s a y f (x) = \sin (x) w i t h x i n d e g r e e .$ $t = \frac{x π}{180} \Rightarrow x = \frac{180 t}{π}$ $\frac{d (\sin x)}{d x} = \frac{d (\sin \frac{180 t}{π})}{\frac{180}{π} d t} = \frac{(\cos \frac{180 t}{π}) \frac{180}{π}}{\frac{180}{π}} = \cos \frac{180 t}{π} = \cos x$ $i t h i n k i n \frac{d y}{d x} w e s e e o n l y f u n c t i o n, i . e .$ $t h e m a t h . r e l a t i o n s h i p b e t w e e n y a n d x,$ $w e {d o n}^{'} t c a r e a b o u t t h e u n i t o f y a n d x .$
Commented by MJS last updated on 26/Apr/20

$I {don}^{'} t think this is true . we plot the function$ $in our coordinate system . if the angle is$ $measured in radiants, the slope of \sin x at$ $x = 0 is 1 . but if the angle is measured in$ $degree the slope cannot be 1 because$ $now the \lim_{h \to 0} \frac{\sin (0 + h) - \sin (0 - h)}{2 h} = \frac{π}{180}$ $\Rightarrow we need a function with f (0) = \frac{π}{180}$
	Answered by TANMAY PANACEA. last updated on 26/Apr/20

	$x d e g r e e = \frac{π}{180} x r a d i a n$ $y = s i n (\frac{π}{180} x)$ $\frac{d y}{d x} = c o s (\frac{π}{180} x) \times \frac{π}{180}$
	Commented by M±th+et+s last updated on 26/Apr/20

	$t h a n k y o u s i r b u t i h a v e a q u e s t i o n$ $w h y s i n (\frac{π}{180} x) = s i n (x)$ $b u t \frac{π}{180} c o s (\frac{π}{180} x) = c o s (x)$ $b e c a u s e w e k n o w \frac{d}{d x} s i n (x) = c o s (x)$
	Answered by M±th+et+s last updated on 27/Apr/20

	$y = s i n (x) y = s i n (\frac{π}{180} x)$ $\frac{π}{180} x = u y = s i n u$ $\frac{d y}{d x} = c o s (u) \frac{d y}{d x} = c o s (\frac{π}{180} x)$
	Commented by M±th+et+s last updated on 27/Apr/20

	$\frac{d y}{d x} s i n (x) = c o s (x)$ $x r a d = \frac{π}{180} x d e g$ $c o s (x) = c o s (\frac{π}{180} x)$ $s o \frac{d y}{d x} s i n (\frac{π}{180} x) = c o s (\frac{π}{180} x)$
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