Question

Using the \lim\limits_{x\to 0} \dfrac{\sin x}{x}=1 and factor identity to prove derivative of sine: (\sin x)^{'} = \cos x

Collected in the board: Derivative

Steven Zheng Steven Zheng posted 1 month ago


Answer

\lim\limits_{h\to 0} \dfrac{\sin (x+h)-\sin x}{h}

= \lim\limits_{h\to 0} \dfrac{2\sin \dfrac{h}{2} \cos (x+\dfrac{h}{2} )}{h}

= \lim\limits_{h\to 0} \dfrac{\sin \dfrac{h}{2}}{ \dfrac{h}{2}} \cos (x+\dfrac{h}{2} )

= \lim\limits_{h\to 0} \dfrac{\sin \dfrac{h}{2}}{ \dfrac{h}{2}} \cdotp \lim\limits_{h\to 0} \cos (x+\dfrac{h}{2} )

=\cos x

Steven Zheng Steven Zheng posted 1 month ago

Scroll to Top