# Derivative by definition of $f(x)= 2x\sin(2x)$

I want to find the derivative of $$f(x)=2x\sin(2x)$$ but I'm having trouble developing the trigonometric identities. Here's what I worked out: \begin{align} f'(x)&=\lim_{h\to0}\frac{2(x+h)\sin(2(x+h))-2x\sin(2x)}{h}\\ &=\lim_{h\to0}\frac{2(x+h)(2\sin(x)\cos(h)+\cos(x)\sin(h))-2x\sin(2x)}{h}\\ &=\lim_{h\to0}\frac{4x\sin(x)\cos(h)+4h\sin(x)\cos(h)+4x\cos(x)\sin(h)+4h\cos(x)\sin(h)-2x\sin(2x)}{h}\end{align}

From this point I separate the expressions and apply special limits but I can't get to the finish line. Is this the right track?

• $\sin(2(x+h))$ is not equal to $2\sin x \cos h + \cos x \sin h$. It's not even equal to $2(\sin x \cos h + \cos x \sin h)$. – TonyK Jan 18 at 13:37
• Isn't $sin(2a)=2sin(a)cos(a)$? And then $sin(a+b)= sin(a)cos(b)+cos(a)sin(b)$? – Jakcjones Jan 18 at 13:49
• Yes...${}{}{}{}{}$ – TonyK Jan 18 at 13:56
• Why is what I did wrong, then? Can you help me understand? – Jakcjones Jan 18 at 14:39
• I mean why can't it be $2(sinxcosh + cosxsinh)?$ – Jakcjones Jan 18 at 14:42

\begin{align} f'(x)&=\lim_{h\to0}\frac{2(x+h)\sin(2x+2h)-2x\sin2x}{h}\\ &=\lim_{h\to0}\left(\frac{2x\sin(2x+2h)-2x\sin2x}{h}+\frac{2h\sin(2x+2h)}{h}\right)\\ &=\lim_{h\to0}\frac{2x(\sin(2x+2h)-\sin2x)}{h}+\lim_{h\to0}\frac{2h\sin(2x+2h)}{h}\\ &=\lim_{h\to0}\frac{2x(2\sin(h)\cos(2x+h))}{h}+\lim_{h\to0}\frac{2\sin(2x+2h)}{1}\\ &=4x\cos2x+2\sin2x \end{align}
• $\sin A-\sin B=2\sin\left(\frac {A-B}{2}\right)\cos\left(\frac {A+B}{2}\right)$. – Thomas Shelby Jan 18 at 14:25
$$\frac{2x\sin2 x-2x_0\sin2 x_0}{x-x_0}=\frac{2x\sin 2x-2x\sin 2x_0+2x\sin2 x_0-2x_0\sin x_0}{x-x_0}=$$
$$=2x\frac{\sin 2x-\sin2 x_0}{x-x_0}+2\sin 2x_0\frac{x-x_0}{x-x_0}\xrightarrow[x\to x_0]{}4x\cos2 x_0+2\sin2 x_0$$