If `y=f(a)` is a differentiable function of a and `a=g(x)` is a differentiable function of x then `y=f(g(x))` is a differentiable function of `x` and `(dy)/(dx)=(dy)/(da)`x`(da)/(dx)`.This rule is also known as CHAIN RULE. Based on the above information find the derivative of functions with respect to `x` in the following questions.
`cos (sqrt(x))`

Options

A. `(cos (sqrt(x)))/(2sqrt(x))`
B. `- (cos (sqrt(x)))/(2sqrt(x))`
C. ` (sin (sqrt(x)))/(2sqrt(x))`
D. `- (sin (sqrt(x)))/(2sqrt(x))`

Explanation

To find the derivative of the function \( y = \cos(\sqrt{x}) \) with respect to \( x \), we will apply the chain rule. Here’s a step-by-step solution:

### Step 1: Identify the outer and inner functions
Let:
- \( u = \sqrt{x} \) (inner function)
- \( y = \cos(u) \) (outer function)

### Step 2: Differentiate the outer function
We need to differentiate \( y \) with respect to \( u \):
\[ \frac{dy}{du} = -\sin(u) \]

### Step 3: Differentiate the inner function
Now, differentiate \( u \) with respect to \( x \):
\[ \frac{du}{dx} = \frac{1}{2\sqrt{x}} \]

### Step 4: Apply the chain rule
Using the chain rule, we have:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = -\sin(u) \cdot \frac{1}{2\sqrt{x}} \]

### Step 5: Substitute back the inner function
Now substitute \( u = \sqrt{x} \) back into the equation:
\[ \frac{dy}{dx} = -\sin(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} \]

### Final Result
Thus, the derivative of \( y = \cos(\sqrt{x}) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = -\frac{\sin(\sqrt{x})}{2\sqrt{x}} \]

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