Can some one show me how to get the derivative of sin2x using the limit definition?

Question

precal · Answer

http://www.wolframalpha.com/input/?i=derivative+of+y%3Dsin%282x%29+by+limit+definition

ash2326 · Answer

Derivative is defined as
$$f'(x)=lim_{h\to 0} \frac{f(x+h)-f(x)}{x+h-h}$$
here f(x) = sin 2x 
$$f'(x)=lim_{h\to 0} \frac{ \sin 2(x+h) - sin 2x}{h}$$Let's expand sin (2x+2h)= sin 2x cos 2h+ sin 2h cos 2x
$$f'(x) =lim_{h\to 0} \frac{ \sin 2x \cos 2h+ \sin 2h cos 2x- \sin 2x}{h}$$
Now cos 2h -->1 as h--->0
so
$$f'(x) =lim_{h\to 0} \frac{ \sin 2x+ \sin 2h \times cos 2x- \sin 2x}{h}$$

$$f'(x) =lim_{h\to 0} \frac{  \sin 2h \times cos 2x}{h}$$
We know
$$\lim_{h \to 0} \frac{sin h}{h}=1$$
so
$$f'(x) =lim_{h\to 0} \frac{2 \times  \sin 2h \times cos 2x}{2h}$$
we get
$$f'(x)= 2cos 2x \lim_{h \to 0} \frac{sin 2h}{2h}$$
Finally
$$f'(x)= 2 \times \cos 2x \times 1$$

anonymous · Answer

Thanks : )