Differentiate ` y= sqrt(x^(2) +5) `w.r. to x

Answer

To differentiate the function \( y = \sqrt{x^2 + 5} \) with respect to \( x \), we will use the chain rule. Here’s a step-by-step solution:

### Step 1: Identify the outer and inner functions
We can express the function as:
- Outer function: \( f(u) = \sqrt{u} \)
- Inner function: \( g(x) = x^2 + 5 \)

Thus, we can rewrite \( y \) as:
\[ y = f(g(x)) = \sqrt{g(x)} = \sqrt{x^2 + 5} \]

**Hint:** Recognize the outer function (the square root) and the inner function (the expression inside the square root).

### Step 2: Differentiate the outer function
The derivative of the outer function \( f(u) = \sqrt{u} \) is:
\[ f'(u) = \frac{1}{2\sqrt{u}} \]

**Hint:** Remember that the derivative of \( \sqrt{u} \) is \( \frac{1}{2\sqrt{u}} \).

### Step 3: Differentiate the inner function
Now, we differentiate the inner function \( g(x) = x^2 + 5 \):
\[ g'(x) = 2x \]

**Hint:** The derivative of \( x^2 \) is \( 2x \), and the derivative of a constant (5) is 0.

### Step 4: Apply the chain rule
Using the chain rule, we have:
\[ \frac{dy}{dx} = f'(g(x)) \cdot g'(x) \]

Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{g(x)}} \cdot g'(x) \]
\[ = \frac{1}{2\sqrt{x^2 + 5}} \cdot 2x \]

### Step 5: Simplify the expression
Now we simplify the expression:
\[ \frac{dy}{dx} = \frac{2x}{2\sqrt{x^2 + 5}} \]
\[ = \frac{x}{\sqrt{x^2 + 5}} \]

### Final Answer
Thus, the derivative of \( y = \sqrt{x^2 + 5} \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{x}{\sqrt{x^2 + 5}} \]

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