Question

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

Answer 1

1.

$f'(\sin x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\sin(x+h) - \sin(x)}{h} = \\ \\ = \lim_{h \to 0} \frac{2 \sin( \frac{x+h - x}{2}) \cdot \cos( \frac{x+h+x}{2}) }{h} = \lim_{h \to 0} \frac{2 \sin( \frac{h}{2}) \cos( \frac{2x+h}{2} ) }{h} = \\ \\ = \lim_{h \to 0} [ \frac{\sin( \frac{h}{2}) }{ \frac{h}{2} } \cdot \cos (\frac{2x+h}{2}) ] = \lim_{h \to 0} [1 \cdot \cos( \frac{2x+h}{2} ) ] =$

$= \cos( \frac{2x}{2}) = \boxed{\cos x}$

2.

$f'(\cos x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\cos(x+h) - \cos(x)}{h} = \\ \\ = \lim_{h \to 0} \frac{-2 \sin ( \frac{x+h+x}{2}) \cdot \sin ( \frac{x+h-x}{2}) }{h} = \lim_{h \to 0} \frac{-2 \sin ( \frac{2x+h}{2}) \cdot \sin ( \frac{h}{2}) }{h} = \\ \\ = \lim_{h \to 0} \frac{-2 \sin ( \frac{2x+h}{2}) }{2} \cdot \frac{sin( \frac{h}{2}) }{ \frac{h}{2} } = \lim_{h \to 0} -\sin( \frac{2x+h}{2}) \cdot 1 =$

$= -\sin( \frac{2x}{2}) = \boxed{\sin x }$

3.

$f'(\tan) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\tan(x+h) - \tan(x)}{h} = \\ \\ = \lim_{h \to 0} \frac{ \frac{\sin(x+h-x)}{\cos(x+h) \cdot \cos(x)} }{h} = \lim_{h \to 0} \frac{ \frac{\sin(h)}{ \frac{\cos(x+h-x) + \cos(x+h+x)}{2} } }{h} =$

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

4.

$f'(\cot) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\cot(x+h) - \cot(x)}{h} = \\ \\ = \lim_{h \to 0} \frac{ \frac{\sin(x - x - h)}{\sin (x+h) \cdot \sin (h)} }{h} = \lim_{h \to 0} \frac{ \frac{\sin(-h) }{ \frac{\cos(x+h-x) - \cos(x+h+x)}{2} } }{h} =$

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

Answer 2

I posted an image instead.