Differentiate the following with respect to 'x' using first principle : `cos(x^(2)+1)`.

Answer

To differentiate the function \( y = \cos(x^2 + 1) \) with respect to \( x \) using the first principle of derivatives, we will follow these steps:

### Step 1: Set up the function and the limit definition
The first principle of derivatives states that:
\[ f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \]
Here, \( f(x) = \cos(x^2 + 1) \).

### Step 2: Calculate \( f(x+h) \)
We need to find \( f(x+h) \):
\[ f(x+h) = \cos((x+h)^2 + 1) \]
Expanding \( (x+h)^2 \):
\[ (x+h)^2 = x^2 + 2xh + h^2 \]
Thus,
\[ f(x+h) = \cos(x^2 + 2xh + h^2 + 1) \]

### Step 3: Substitute into the limit definition
Now we substitute \( f(x+h) \) and \( f(x) \) into the limit definition:
\[ f'(x) = \lim_{h \to 0} \frac{\cos(x^2 + 2xh + h^2 + 1) - \cos(x^2 + 1)}{h} \]

### Step 4: Use the cosine subtraction formula
We can use the identity \( \cos A - \cos B = -2 \sin\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \):
Let \( A = x^2 + 2xh + h^2 + 1 \) and \( B = x^2 + 1 \).
Then,
\[ A - B = 2xh + h^2 \]
\[ A + B = 2(x^2 + 1) + 2xh + h^2 \]
Substituting these into the limit gives:
\[ f'(x) = \lim_{h \to 0} \frac{-2 \sin\left(x^2 + 1 + xh + \frac{h^2}{2}\right) \sin\left(xh + \frac{h^2}{2}\right)}{h} \]

### Step 5: Simplify the limit
As \( h \to 0 \), \( \sin\left(xh + \frac{h^2}{2}\right) \approx xh \) (using the small angle approximation \( \sin u \approx u \)):
\[ f'(x) = \lim_{h \to 0} -2 \sin(x^2 + 1 + 0) \cdot \frac{xh}{h} \]
This simplifies to:
\[ f'(x) = -2 \sin(x^2 + 1) \cdot x \]

### Final Result
Thus, the derivative of \( y = \cos(x^2 + 1) \) is:
\[ f'(x) = -2x \sin(x^2 + 1) \]