How do you differentiate y = sin 2x – cos 2x using rule?

Hint: We are given an equation y = sin 2x – cos 2x. We are asked to find its derivative. We will learn about the chain rule. We will use the chain rule which is given as \[d\left[ f\left( g\left( x \right) \right) \right]={{f}^{'}}\left( g\left( x \right) \right).{{g}^{'}}\left( x \right)\] along with that we will use that the derivative of difference in the difference of the derivative that is d(A – B) = d(A) – d(B). We will use that d(cos x) = – sin x, d(sin x) = cos x and we need \[d\left( {{x}^{n}} \right)=n.{{x}^{n-1}}\] to find the value of our given problem.

Complete step by step answer:
We have y = sin 2x – cos 2x, we have to find the derivative of this function. Before we start, we will learn things about derivatives and we will learn some rules. First, we see that if we find the derivative of the difference of two functions then the derivative of them is the same as differentiating those functions separately and then finding the difference. If A(x) and B(x) are two functions, then
\[\dfrac{d\left( A\left( x \right)-B\left( x \right) \right)}{dx}=\dfrac{d\left( A\left( x \right) \right)}{dx}-\dfrac{d\left( B\left( x \right) \right)}{dx}\]
Now, we learn what the chain rule is. If any function is made by the composition of two functions, then for these we use the chain rule. If we have, \[\left( fog\left( x \right) \right)=f\left[ g\left( x \right) \right],\] then \[\dfrac{f\left[ f\left( g\left( x \right) \right) \right]}{dx}={{f}^{'}}\left[ g\left( x \right) \right].{{g}^{'}}\left( x \right).\] We will use these to solve our problem, so we have as y = sin 2x – cos 2x. On differentiating, we get,
\[\dfrac{dy}{dx}=\dfrac{d\left( \sin 2x-\cos 2x \right)}{dx}\]
On simplifying using \[\dfrac{d\left( A-B \right)}{dx}=\dfrac{dA}{dx}-\dfrac{dB}{dx},\] we get
\[\Rightarrow \dfrac{d\left( \sin \left( 2x \right) \right)}{dx}-\dfrac{d\left( \cos \left( 2x \right) \right)}{dx}\]
Now, we use the chain rule on these two to solve, we will find the derivative one by one. As we can see sin 2x is formed by the composition of sin x and 2x, so let f(x) = sin x and g(x) = 2x. So, we get,
\[\dfrac{d\left( \sin \left( 2x \right) \right)}{dx}=\cos \left( 2x \right).2\]
[As f’(g(x)) = cos 2x and g’(x) = 2]
And similarly,
\[\dfrac{d\left( \cos \left( 2x \right) \right)}{dx}=-\sin \left( 2x \right).2\]
[As cos x derivative is – sin x and 2x derivative is 2]
So, using these above, we get,
\[\Rightarrow \dfrac{dy}{dx}=2\cos \left( 2x \right)-\left[ -2\sin \left( 2x \right) \right]\]
\[\Rightarrow \dfrac{dy}{dx}=2\cos \left( 2x \right)+2\sin \left( 2x \right)\]
Taking 2 common, we get,
\[\Rightarrow \dfrac{dy}{dx}=2\left[ \cos \left( 2x \right)+\sin \left( 2x \right) \right]\].

Note:
While simplifying the fraction, remember we can add like terms only. While calculating the derivative, we need to be very careful while picking the function as if our function is comprised of more than one function, then its derivative will be different and that is usually like \[\dfrac{d\left( {{e}^{2x}} \right)}{dx}\ne {{e}^{2x}}\] as it was made up by the composition of \[{{e}^{x}}\] and 2x. Also, while differentiating we need to carefully use \[d\left( {{x}^{n}} \right)=n.{{x}^{n-1}},\] a simple error that may happen while calculating.