Question

Find derivative using limit definition: sin(alpha x) [alpha looks like a sideways swirl?]

Answer 1

\[f(x) := \sin(\alpha x)\]
\[f'(x) =\lim_{t \rightarrow 0} \frac{\sin(\alpha (x + t)) - \sin(\alpha x)}{t} =\lim_{t \rightarrow 0} \frac{\sin(\alpha x)\cos(\alpha t) + \cos(\alpha x)\sin(\alpha t)- \sin(\alpha x)}{t} \]
\[\sin(\alpha x)\lim_{t \rightarrow 0} \frac{\cos(\alpha t)- 1}{t} + \cos(\alpha x)\lim_{t \rightarrow 0} \frac{\sin(\alpha t)}{t} = \sin(\alpha x)\lim_{t \rightarrow 0} \frac{(\cos(\alpha t)- 1)(\cos(\alpha t)+ 1)}{t(\cos(\alpha t)+ 1)} + \cos(\alpha x)\alpha=\]
\[\sin(\alpha x)\lim_{t \rightarrow 0} \frac{\cos^2(\alpha t) - 1}{t(\cos(\alpha t)+ 1)} + \cos(\alpha x)\alpha= \sin(\alpha x)\lim_{t \rightarrow 0} \frac{-\sin^2(\alpha t) }{t(\cos(\alpha t)+ 1)} + \cos(\alpha x)\alpha=\]
\[\sin(\alpha x)*\lim_{t \rightarrow 0}( \sin(\alpha t) * \frac{-\sin(\alpha t)}{t}*\frac{1}{\cos(\alpha t +1)} )+ \cos(\alpha x)\alpha= \]
\[\sin(\alpha x)*0*\alpha *1 + \cos(\alpha x)\alpha = \cos(\alpha x)\alpha\]

Answer 2

woah! o.O thanks

Answer 3

hmmmmmm it didn't show it right, the second line is not complete

Answer 4

i can still follow what you did

Answer 5

Great :) tell me if something is unclear