To differentiate the function [tex]\( f(x) = x \cos x - \sin x \)[/tex], we will apply the rules of differentiation step-by-step.
1. Differentiate [tex]\( x \cos x \)[/tex]
Here, we need to use the product rule for differentiation. Recall that the product rule states:
[tex]\[
\frac{d}{dx} [u(x) v(x)] = u'(x) v(x) + u(x) v'(x)
\][/tex]
For [tex]\( u(x) = x \)[/tex] and [tex]\( v(x) = \cos x \)[/tex]:
[tex]\[
u'(x) = 1 \quad \text{and} \quad v'(x) = -\sin x
\][/tex]
Applying the product rule:
[tex]\[
\frac{d}{dx} [x \cos x] = (1) \cos x + x (-\sin x) = \cos x - x \sin x
\][/tex]
2. Differentiate [tex]\(-\sin x \)[/tex]
The derivative of [tex]\(-\sin x\)[/tex] is:
[tex]\[
\frac{d}{dx} [-\sin x] = -\cos x
\][/tex]
3. Combine the results
Now, we combine the results of the derivatives obtained from steps 1 and 2:
[tex]\[
\frac{d}{dx} \left[ x \cos x - \sin x \right] = \left( \cos x - x \sin x \right) - \left( \cos x \right)
\][/tex]
Simplifying, the [tex]\(\cos x\)[/tex] terms cancel out:
[tex]\[
\cos x - x \sin x - \cos x = -x \sin x
\][/tex]
Therefore, the derivative of [tex]\( f(x) = x \cos x - \sin x \)[/tex] is:
[tex]\[
f'(x) = -x \sin x
\][/tex]
