Question

find the derivative of y=cos(2x) using limit definition.

Answer 1

\[\lim_{h\to 0}\frac{\cos{(2(x+h))}-\cos{(2x)}}{h}\]

Answer 2

\[\frac{\cos{(2x+2h)-\cos{(2x)}}}{h}\]
\[\frac{\cos(2x)\cos(2h)-\sin(2x)\sin(2h)-\cos(2x)}{h}\]
\[\frac{\cos(2x)(\cos(2h)-1)-\sin(2x)\sin(2h)}{h}\]
\[\frac{\cos(2x)(\cos(2h)-1)}{h}-\frac{\sin(2x)\sin(2h)}{h}\]

\[\frac{\cos(2x)(\cos^2(h)-\sin^2(h)-1)}{h}-\frac{\sin(2x)(2)\sin(h)\cos(h)}{h}\]
\[\frac{-\cos(2x)(-\cos^2(h)+\sin^2(h)+1)}{h}-\frac{\sin(2x)(2)\sin(h)\cos(h)}{h}\]
Notice that \[ -\cos^2(x)+1=\sin^2(x)\]
\[\lim_{h\to 0}\frac{-\cos(2x)(2\sin^2(h))}{h}-\lim_{h\to 0}\frac{\sin(2x)(2)\sin(h)\cos(h)}{h}\]
\[\lim_{h\to 0} \frac{\sin(h)}{h}=1\]
so
\[\lim_{h\to 0}-\cos(2x)(2\sin(h))-\lim_{h\to 0}\sin(2x)(2)\cos(h)\]
as h goes to 0, cos(h) = 1 and sin(h)=0
so
\[\lim_{h\to 0}-\cos(2x)(2)(0)-\lim_{h\to 0}\sin(2x)(2)(1)\]
\[0-\sin(2x)(2)=-2\sin(2x)\]