Differentiate $ \sin 5x\cos 3x $ with respect to x.

Hint: We can find the differentiation of $ \sin 5x\cos 3x $ using product rule. Product rule is a rule to differentiate problems where one function is multiplied by another function. We cannot directly differentiate a product of functions so we separate and differentiate the functions according to the product rule.
Formula used:
According to the product rule, $ \dfrac{d}{{dx}}\left( {uv} \right) = u\left( {\dfrac{{dv}}{{dx}}} \right) + v\left( {\dfrac{{du}}{{dx}}} \right) $

Complete step-by-step answer:
We are given to differentiate $ \sin 5x\cos 3x $ with respect to x.
Here on comparing $ \sin 5x\cos 3x $ with $ uv $ , we get the value of u as $ \sin 5x $ and the value of v as $ \cos 3x $
 Differentiation $ \dfrac{d}{{dx}}\left( {uv} \right) = u\left( {\dfrac{{dv}}{{dx}}} \right) + v\left( {\dfrac{{du}}{{dx}}} \right) $
In the same way the differentiation of $ \sin 5x\cos 3x $ with respect to x is $ \dfrac{d}{{dx}}\left( {\sin 5x\cos 3x} \right) $ .
 $ \dfrac{d}{{dx}}\left( {\sin 5x\cos 3x} \right) = \left( {\sin 5x} \right)\dfrac{d}{{dx}}\left( {\cos 3x} \right) + \left( {\cos 3x} \right)\dfrac{d}{{dx}}\left( {\sin 5x} \right) $
We know that differentiation of cosine function is negative sine function and differentiation of sine function is positive cosine function.
 $ \dfrac{d}{{dx}}\cos x = - \sin x,\dfrac{d}{{dx}}\sin x = \cos x $
But here in the place of x we have 3x for cosine and in the place of x we have 5x for sine.
So we have to consider 3x and 5x as separate functions.
 $ \dfrac{d}{{dx}}\cos f\left( x \right) = - \sin f\left( x \right) \times \dfrac{d}{{dx}}f\left( x \right) $ and $ \dfrac{d}{{dx}}\sin f\left( x \right) = \cos f\left( x \right) \times \dfrac{d}{{dx}}f\left( x \right) $
Here $ f\left( x \right) $ is 3x for cosine and $ f\left( x \right) $ is 5x for sine.
Differentiation of 3x is 3 and differentiation of 5x is 5.
Therefore, $ \dfrac{d}{{dx}}\cos 3x = - \sin 3x \times 3 = - 3\sin 3x $ and $ \dfrac{d}{{dx}}\sin 5x = \cos 5x \times 5 = 5\cos 5x $
On substituting the above obtained values in $ \dfrac{d}{{dx}}\left( {\sin 5x\cos 3x} \right) $ , we get
 $ \Rightarrow\dfrac{d}{{dx}}\left( {\sin 5x\cos 3x} \right) = \sin 5x\left( { - 3\sin 3x} \right) + \cos 3x\left( {5\cos 5x} \right) $
 $ \Rightarrow\dfrac{d}{{dx}}\left( {\sin 5x\cos 3x} \right) = - 3\sin 5x\sin 3x + 5\cos 3x\cos 5x = 5\cos 3x\cos 5x - 3\sin 3x\sin 5x $
Therefore, differentiation of $ \sin 5x\cos 3x $ with respect to x is $ 5\cos 3x\cos 5x - 3\sin 3x\sin 5x $

So, the correct answer is “ $ 5\cos 3x\cos 5x - 3\sin 3x\sin 5x $ ”.

Note: We can also further simplify the obtained differentiation values by using the formulas $ \sin A\sin B $ and $ \cos A\cos B $ where A will be 3x and B will be 5x. The result we get after using these formulas will only be in terms of cosine function. Do not confuse between differentiation and integration. For example, differentiation of sine gives positive cosine whereas integration of sine gives negative cosine. As we can except for the sign, the whole result is the same. So be careful with the results of differentiations and integration.