Differentiate the following with respect to 'x' using first principle : `xcosx`

Answer

To differentiate the function \( f(x) = x \cos x \) using the first principle of derivatives, we follow these steps:

### Step 1: Define the function and the increment
We start with the function:
\[ f(x) = x \cos x \]
We need to find \( f'(x) \) using the first principle, which states:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Calculate \( f(x+h) \)
Next, we calculate \( f(x+h) \):
\[ f(x+h) = (x+h) \cos(x+h) \]

### Step 3: Substitute into the limit formula
Now we substitute \( f(x+h) \) and \( f(x) \) into the limit formula:
\[ f'(x) = \lim_{h \to 0} \frac{(x+h) \cos(x+h) - x \cos x}{h} \]

### Step 4: Expand the expression
Expanding the expression in the numerator:
\[ = \lim_{h \to 0} \frac{x \cos(x+h) + h \cos(x+h) - x \cos x}{h} \]
This can be rearranged as:
\[ = \lim_{h \to 0} \left( \frac{x (\cos(x+h) - \cos x)}{h} + \cos(x+h) \right) \]

### Step 5: Apply the limit
Now we can evaluate the limit. The first term involves the derivative of cosine:
\[ \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h} = -\sin x \]
Thus, we have:
\[ = x(-\sin x) + \lim_{h \to 0} \cos(x+h) \]
As \( h \to 0 \), \( \cos(x+h) \to \cos x \):
\[ = -x \sin x + \cos x \]

### Step 6: Final result
Therefore, the derivative of \( f(x) = x \cos x \) is:
\[ f'(x) = \cos x - x \sin x \]