Using definition, differentiate w.r.t. 'x' `f(x)=cos^(2)x`

Answer

To differentiate the function \( f(x) = \cos^2 x \) with respect to \( x \) using the definition of the derivative, we follow these steps:

### Step 1: Define the function
Let \( y = f(x) = \cos^2 x \).

### Step 2: Write the expression for \( y + \Delta y \)
When \( x \) increases by \( \Delta x \), the new value of \( y \) becomes:
\[ y + \Delta y = \cos^2(x + \Delta x) \]

### Step 3: Set up the difference quotient
The difference in \( y \) can be expressed as:
\[ \Delta y = \cos^2(x + \Delta x) - \cos^2 x \]
Thus, we can write the difference quotient:
\[ \frac{\Delta y}{\Delta x} = \frac{\cos^2(x + \Delta x) - \cos^2 x}{\Delta x} \]

### Step 4: Use the identity for the difference of squares
We can use the identity \( a^2 - b^2 = (a - b)(a + b) \) to rewrite \( \Delta y \):
\[ \Delta y = \cos^2(x + \Delta x) - \cos^2 x = (\cos(x + \Delta x) - \cos x)(\cos(x + \Delta x) + \cos x) \]

### Step 5: Substitute into the difference quotient
Substituting this back into our difference quotient gives:
\[ \frac{\Delta y}{\Delta x} = \frac{(\cos(x + \Delta x) - \cos x)(\cos(x + \Delta x) + \cos x)}{\Delta x} \]

### Step 6: Find the limit as \( \Delta x \to 0 \)
Now, we need to find the limit:
\[ \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} \]
We know that:
\[ \lim_{\Delta x \to 0} \frac{\cos(x + \Delta x) - \cos x}{\Delta x} = -\sin x \]
Thus, we can express the limit as:
\[ \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \left( -\sin x (\cos(x + \Delta x) + \cos x) \right) \]
As \( \Delta x \to 0 \), \( \cos(x + \Delta x) \to \cos x \), so:
\[ \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = -\sin x (2\cos x) = -2\sin x \cos x \]

### Step 7: Final result
Thus, the derivative of \( f(x) = \cos^2 x \) with respect to \( x \) is:
\[ \frac{dy}{dx} = -2\sin x \cos x \]