Find the derivative of the following functions from first principles :
`sin(x+1)`

Answer

To find the derivative of the function \( f(x) = \sin(x + 1) \) from first principles, we will use the definition of the derivative. The derivative \( f'(x) \) is given by:

\[ f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \]

### Step 1: Substitute the function into the formula

First, we need to find \( f(x + h) \):

\[ f(x + h) = \sin((x + h) + 1) = \sin(x + h + 1) \]

Now, we can substitute \( f(x + h) \) and \( f(x) \) into the derivative formula:

\[ f'(x) = \lim_{h \to 0} \frac{\sin(x + h + 1) - \sin(x + 1)}{h} \]

### Step 2: Use the sine subtraction formula

Next, we apply the sine subtraction formula, which states that \( \sin A - \sin B = 2 \cos\left(\frac{A + B}{2}\right) \sin\left(\frac{A - B}{2}\right) \):

Let \( A = x + h + 1 \) and \( B = x + 1 \). Then:

\[ f'(x) = \lim_{h \to 0} \frac{2 \cos\left(\frac{(x + h + 1) + (x + 1)}{2}\right) \sin\left(\frac{(x + h + 1) - (x + 1)}{2}\right)}{h} \]

This simplifies to:

\[ = \lim_{h \to 0} \frac{2 \cos\left(x + 1 + \frac{h}{2}\right) \sin\left(\frac{h}{2}\right)}{h} \]

### Step 3: Simplify the expression

Now, we can rewrite \( \sin\left(\frac{h}{2}\right) \) as follows:

\[ \sin\left(\frac{h}{2}\right) \approx \frac{h}{2} \text{ as } h \to 0 \]

Thus, we can substitute this approximation into our limit:

\[ f'(x) = \lim_{h \to 0} \frac{2 \cos\left(x + 1 + \frac{h}{2}\right) \cdot \frac{h}{2}}{h} \]

This simplifies to:

\[ = \lim_{h \to 0} \cos\left(x + 1 + \frac{h}{2}\right) \]

### Step 4: Evaluate the limit

As \( h \to 0 \), \( \frac{h}{2} \to 0 \), so we have:

\[ f'(x) = \cos(x + 1) \]

### Final Result

Thus, the derivative of \( f(x) = \sin(x + 1) \) is:

\[ f'(x) = \cos(x + 1) \]

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