Differentiate the following functions w.r.t. x.
`(x^3 + sin x)^5`

Answer

To differentiate the function \( (x^3 + \sin x)^5 \) with respect to \( x \), we will use the chain rule of differentiation. Here’s a step-by-step solution:

### Step 1: Identify the outer and inner functions
Let:
\[ y = (x^3 + \sin x)^5 \]
Here, the outer function is \( u^5 \) and the inner function is \( u = x^3 + \sin x \).

### Step 2: Apply the chain rule
According to the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]

### Step 3: Differentiate the outer function
First, we differentiate the outer function \( y = u^5 \):
\[ \frac{dy}{du} = 5u^4 \]

### Step 4: Differentiate the inner function
Next, we differentiate the inner function \( u = x^3 + \sin x \):
\[ \frac{du}{dx} = \frac{d}{dx}(x^3) + \frac{d}{dx}(\sin x) = 3x^2 + \cos x \]

### Step 5: Combine the results
Now, we substitute \( u \) back into the derivative:
\[ \frac{dy}{dx} = 5u^4 \cdot (3x^2 + \cos x) \]
Replacing \( u \) with \( x^3 + \sin x \):
\[ \frac{dy}{dx} = 5(x^3 + \sin x)^4 \cdot (3x^2 + \cos x) \]

### Final Answer
Thus, the derivative of the function \( (x^3 + \sin x)^5 \) with respect to \( x \) is:
\[ \frac{dy}{dx} = 5(x^3 + \sin x)^4 (3x^2 + \cos x) \]

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