Differentiate sin `sqrtx+cos(x^2)` w.r.t. x

Answer

To differentiate the function \( f(x) = \sin(\sqrt{x}) + \cos(x^2) \) with respect to \( x \), we will use the chain rule and the basic differentiation rules. Here’s the step-by-step solution:

### Step 1: Differentiate \( \sin(\sqrt{x}) \)

Using the chain rule:
\[ \frac{d}{dx} \sin(u) = \cos(u) \cdot \frac{du}{dx} \]
where \( u = \sqrt{x} \).

Now, we need to find \( \frac{du}{dx} \):
\[ u = \sqrt{x} = x^{1/2} \]
\[ \frac{du}{dx} = \frac{1}{2} x^{-1/2} = \frac{1}{2\sqrt{x}} \]

Now, applying the chain rule:
\[ \frac{d}{dx} \sin(\sqrt{x}) = \cos(\sqrt{x}) \cdot \frac{1}{2\sqrt{x}} = \frac{\cos(\sqrt{x})}{2\sqrt{x}} \]

### Step 2: Differentiate \( \cos(x^2) \)

Again, using the chain rule:
\[ \frac{d}{dx} \cos(v) = -\sin(v) \cdot \frac{dv}{dx} \]
where \( v = x^2 \).

Now, we find \( \frac{dv}{dx} \):
\[ v = x^2 \]
\[ \frac{dv}{dx} = 2x \]

Now, applying the chain rule:
\[ \frac{d}{dx} \cos(x^2) = -\sin(x^2) \cdot 2x = -2x \sin(x^2) \]

### Step 3: Combine the results

Now, we combine the derivatives from Step 1 and Step 2:
\[ \frac{d}{dx} f(x) = \frac{\cos(\sqrt{x})}{2\sqrt{x}} - 2x \sin(x^2) \]

### Final Answer
Thus, the derivative of \( f(x) = \sin(\sqrt{x}) + \cos(x^2) \) with respect to \( x \) is:
\[ f'(x) = \frac{\cos(\sqrt{x})}{2\sqrt{x}} - 2x \sin(x^2) \]

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