Question

Find the derivative of the transcendental function. $$ y=x \cos x+\sin x $$

Short answer

The derivative of the function \(y=x \cos x+\sin x\) is \(2 \cos x - x \sin x\)

Step-by-step solution

Identify the individual functions

Firstly, differentiate each function individually. The derivative of \(x\) is \(1\), the derivative of \(\sin x\) is \(\cos x\), and the derivative of \(\cos x\) is \(-\sin x\)

Apply the Product rule

The expression \(x \cos x\) is a product of two functions, hence the product rule of differentiation applies. The product rule is expressed as (uv)' = u'v + uv', where \(u\) and \(v\) are functions of \(x\). Here, \(u = x\), \(v = \cos x\), \(u' = 1\) and \(v' = -\sin x\). Applying the product rule we find that the derivative \(x \cos x\) is \(1. \cos x + x. -\sin x\) which simplifies to become \(\cos x - x \sin x\)

Apply the Sum rule

Next, apply the sum rule of differentiation to the original expression \(x \cos x + \sin x\), where the sum rule (u+v)' = u' + v' is used. Here, \(u' = \cos x - x \sin x\) (the derivative of \(x \cos x\)) and \(v' = \cos x\) (the derivative of \(\sin x\)). Adding these together, we find that the derivative of \(x \cos x + \sin x\) is \(\cos x - x \sin x + \cos x\). Simplifying by combining like terms, we have \(2 \cos x - x \sin x\).