Question

How do you find the derivative of \[2\sin x\cos x\]?

Answer 1

Hint: In the given question, we have been given a trigonometric expression. It is a product of two trigonometric functions and a constant. We have to find the derivative of this trigonometric expression. To do that, we apply the product rule of differentiation on the trigonometric functions.

Formula Used:
We are going to use the formula of product rule of differentiation, which is:
\[{\left( {uv} \right)^\prime } = u'v + uv'\]

Complete step-by-step answer:
The given trigonometric expression is \[2\sin x\cos x\].
We are going to use the formula of product rule of differentiation, which is:
\[{\left( {uv} \right)^\prime } = u'v + uv'\]
Here, \[u = \sin x\] and \[v = \cos x\]
Substituting the values into the formula, we get,
\[2\sin x\cos x = 2\left( {{{\left( {\sin x} \right)}^\prime }\cos x + {{\left( {\cos x} \right)}^\prime }\sin x} \right)\]
Now, \[{\left( {\sin x} \right)^\prime } = \cos x\] and \[{\left( {\cos x} \right)^\prime } = - \sin x\]
Hence,
\[ = 2\left( {\cos x.\cos x + \left( { - \sin x} \right).\sin x} \right)\]
Multiplying them,
\[ = 2\left( {{{\cos }^2}x - {{\sin }^2}x} \right)\]
Now, we know that
\[\cos 2x = {\cos ^2}x - {\sin ^2}x\]
Hence,
\[ = 2\cos 2x\]
Thus, the derivative of \[2\sin x\cos x\] is \[2\cos 2x\].

Additional Information:
The given expression \[2\sin x\cos x\] can be also represented as:
\[\sin 2x\]

Note: In the given question, we had been given a trigonometric expression. It was a product of two trigonometric functions. We had to find the derivative of this trigonometric expression. We solved this question by applying the product rule of differentiation on the trigonometric functions. Whenever there are two functions, we always apply this rule for finding the derivative of the functions.