Find the derivative of `cos (x^(2) + 3)` using first principle of derivatives.

Answer

To find the derivative of \( f(x) = \cos(x^2 + 3) \) using the first principle of derivatives, we will follow these steps:

### Step 1: Write the definition of the derivative
The derivative of a function \( f(x) \) using the first principle is given by:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]

### Step 2: Substitute \( f(x) \) into the formula
For our function \( f(x) = \cos(x^2 + 3) \), we need to compute \( f(x+h) \):
\[ f(x+h) = \cos((x+h)^2 + 3) = \cos(x^2 + 2xh + h^2 + 3) \]
Now, substituting into the derivative formula:
\[ f'(x) = \lim_{h \to 0} \frac{\cos(x^2 + 2xh + h^2 + 3) - \cos(x^2 + 3)}{h} \]

### Step 3: Use the cosine difference identity
We can use the identity for the difference of cosines:
\[ \cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right) \]
Let \( A = x^2 + 2xh + h^2 + 3 \) and \( B = x^2 + 3 \). Then:
\[ A - B = 2xh + h^2 \]
\[ A + B = (x^2 + 2xh + h^2 + 3) + (x^2 + 3) = 2x^2 + 2xh + h^2 + 6 \]
Now substituting into the limit:
\[ f'(x) = \lim_{h \to 0} \frac{-2 \sin\left(\frac{2x^2 + 2xh + h^2 + 6}{2}\right) \sin\left(\frac{2xh + h^2}{2}\right)}{h} \]

### Step 4: Simplify the expression
The expression simplifies to:
\[ f'(x) = \lim_{h \to 0} -2 \sin\left(x^2 + xh + \frac{h^2}{2} + 3\right) \cdot \frac{\sin\left(xh + \frac{h^2}{2}\right)}{h} \]

### Step 5: Apply the limit
As \( h \to 0 \), \( \sin\left(xh + \frac{h^2}{2}\right) \approx xh + \frac{h^2}{2} \) and we can use the limit property:
\[ \lim_{h \to 0} \frac{\sin(kh)}{h} = k \quad \text{for small } h \]
Thus:
\[ \lim_{h \to 0} \frac{\sin\left(xh + \frac{h^2}{2}\right)}{h} = x \]

### Step 6: Final expression
Now substituting back into our limit:
\[ f'(x) = -2 \sin(x^2 + 3) \cdot x \]

### Final Result
Thus, the derivative of \( f(x) = \cos(x^2 + 3) \) is:
\[ f'(x) = -2x \sin(x^2 + 3) \]

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