# How do you differentiate  y= sin2x-cos2x using the chain rule?

##### 1 Answer
Jan 31, 2016

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \cos 2 x + 2 \sin 2 x$

#### Explanation:

The chain rule basically says that the derivative of a composite function (like $\sin 2 x$) is the derivative of the main function times the derivative of the interior function. In the case of $\sin 2 x$, we have the derivative of the main function ($\sin 2 x$ ) as $\cos 2 x$, and the derivative of the interior function ($2 x$) as $2$. So, by the chain rule, the derivative of $\sin 2 x$ is $2 \cos 2 x$.

The same logic applies to $\cos 2 x$. The derivative of cosine is negative sine ($- \sin$). so we begin with $- \sin 2 x$. Then, because of the chain rule, we multiply this by the derivative of $2 x$, which is, of course, $2$. Therefore, the derivative of $\cos 2 x$ is $- 2 \sin 2 x$.

Finally we put it all together. We can express $\frac{\mathrm{dy}}{\mathrm{dx}} \left(\sin 2 x - \cos 2 x\right)$ as $\frac{\mathrm{dy}}{\mathrm{dx}} \left(\sin 2 x\right) - \frac{\mathrm{dy}}{\mathrm{dx}} \left(\cos 2 x\right)$. Because we found the derivatives of these guys earlier, all we do is substitute:

$\frac{\mathrm{dy}}{\mathrm{dx}} \left(\sin 2 x\right) = 2 \cos 2 x$ and $\frac{\mathrm{dy}}{\mathrm{dx}} \left(\cos 2 x\right) = - 2 \sin 2 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = 2 \cos 2 x - \left(- 2 \sin 2 x\right) = 2 \cos 2 x + 2 \sin 2 x$