Question

Explain the derivation behind the derivative of sin(x) i.e. prove f'(sin(x)) = cos(x)

How about cos(x) and tan(x)?

Answer 1

I posted an image instead.

Answer 2

1.


`f'(\sin x) = \lim_(h \to 0) (f(x+h) - f(x))/(h) = \lim_(h \to 0) (\sin(x+h) - \sin(x))/(h) = \\ \\ = \lim_(h \to 0) (2 \sin( (x+h - x)/(2)) \cdot \cos( (x+h+x)/(2)) )/(h) = \lim_(h \to 0) (2 \sin( (h)/(2)) \cos( (2x+h)/(2) ) )/(h) = \\ \\ = \lim_(h \to 0) [ (\sin( (h)/(2)) )/( (h)/(2) ) \cdot \cos ((2x+h)/(2)) ] = \lim_(h \to 0) [1 \cdot \cos( (2x+h)/(2) ) ] =`


`= \cos( (2x)/(2)) = \boxed{\cos x}`

2.


`f'(\cos x) = \lim_(h \to 0) (f(x+h) - f(x))/(h) = \lim_(h \to 0) (\cos(x+h) - \cos(x))/(h) = \\ \\ = \lim_(h \to 0) (-2 \sin ( (x+h+x)/(2)) \cdot \sin ( (x+h-x)/(2)) )/(h) = \lim_(h \to 0) (-2 \sin ( (2x+h)/(2)) \cdot \sin ( (h)/(2)) )/(h) = \\ \\ = \lim_(h \to 0) (-2 \sin ( (2x+h)/(2)) )/(2) \cdot (sin( (h)/(2)) )/( (h)/(2) ) = \lim_(h \to 0) -\sin( (2x+h)/(2)) \cdot 1 =`


`= -\sin( (2x)/(2)) = \boxed{\sin x }`

3.


`f'(\tan) = \lim_(h \to 0) (f(x+h) - f(x))/(h) = \lim_(h \to 0) (\tan(x+h) - \tan(x))/(h) = \\ \\ = \lim_(h \to 0) ( (\sin(x+h-x))/(\cos(x+h) \cdot \cos(x)) )/(h) = \lim_(h \to 0) ( (\sin(h))/( (\cos(x+h-x) + \cos(x+h+x))/(2) ) )/(h) =`


`= \lim_(h \to 0) ( (\sin(h))/(\cos(h) + \cos(2x+h)) )/( (1)/(2)h ) = \lim_(h \to 0) (\sin(h))/( (1)/(2)h \cdot [\cos(h) + \cos(2x+h)] ) = \\ \\ = \lim_(h \to 0) (\sin(h))/(h) \cdot (1)/( (1)/(2) \cdot (\cos(h) + cos(2x+h) ) = 1 \cdot (1)/( (1)/(2) \cdot (1+ cos(2x) ) = (2)/(1 + 2 \cos^(2) - 1 ) = \\ \\ = (2)/(2 \cos^(2) x) = \boxed{ (1)/(\cos^(2)x) }`

4.


`f'(\cot) = \lim_(h \to 0) (f(x+h) - f(x))/(h) = \lim_(h \to 0) (\cot(x+h) - \cot(x))/(h) = \\ \\ = \lim_(h \to 0) ( (\sin(x - x - h))/(\sin (x+h) \cdot \sin (h)) )/(h) = \lim_(h \to 0) ( (\sin(-h) )/( (\cos(x+h-x) - \cos(x+h+x))/(2) ) )/(h) =`


`= \lim_(h \to 0) ( (-\sin(h))/(\cos(h) - \cos(2x+h)) )/( (1)/(2)h ) = \lim_(h \to 0) ( - \sin(h))/( (1)/(2)h \cdot [\cos(h) - \cos(2x+h)] ) = \\ \\ = \lim_(h \to 0) (- \sin (h))/(h) \cdot (1)/( (1)/(2) \cdot [\cos(h) - \cos(2x+h)] ) = -1 \cdot (2)/(1 - cos(2x)) = \\ \\ = - (2)/(1 -1 + 2 \sin^(2)x) = - (2)/(2 \sin^(2) x) = \boxed{- (1)/(\sin^(2) x) }`