Differentiate the following w.r.t. x or t or u as the case may be:
2. `(x^(100) + 2x^(50) - 3) (7x^(8) + 20x + 5)`

Answer

To differentiate the function \( y = (x^{100} + 2x^{50} - 3)(7x^{8} + 20x + 5) \) with respect to \( x \), we will use the product rule. The product rule states that if you have two functions \( u \) and \( v \), then the derivative \( \frac{dy}{dx} \) is given by:

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

### Step 1: Identify \( u \) and \( v \)

Let:
- \( u = x^{100} + 2x^{50} - 3 \)
- \( v = 7x^{8} + 20x + 5 \)

### Step 2: Differentiate \( u \) and \( v \)

Now we need to find the derivatives \( \frac{du}{dx} \) and \( \frac{dv}{dx} \).

#### Differentiate \( u \):
\[ \frac{du}{dx} = \frac{d}{dx}(x^{100}) + \frac{d}{dx}(2x^{50}) + \frac{d}{dx}(-3) \]
Using the power rule:
\[ \frac{du}{dx} = 100x^{99} + 100x^{49} + 0 = 100x^{99} + 100x^{49} \]

#### Differentiate \( v \):
\[ \frac{dv}{dx} = \frac{d}{dx}(7x^{8}) + \frac{d}{dx}(20x) + \frac{d}{dx}(5) \]
Using the power rule:
\[ \frac{dv}{dx} = 56x^{7} + 20 + 0 = 56x^{7} + 20 \]

### Step 3: Apply the Product Rule

Now we can apply the product rule:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \):
\[ \frac{dy}{dx} = (x^{100} + 2x^{50} - 3)(56x^{7} + 20) + (7x^{8} + 20x + 5)(100x^{99} + 100x^{49}) \]

### Step 4: Simplify the Expression

Now we simplify the expression:
1. **First Term**:
\[ (x^{100} + 2x^{50} - 3)(56x^{7} + 20) = 56x^{107} + 40x^{50} - 168 + 112x^{57} - 3 \cdot 20 \]
2. **Second Term**:
\[ (7x^{8} + 20x + 5)(100x^{99} + 100x^{49}) = 700x^{107} + 700x^{57} + 500x^{49} + 2000x^{49} \]

Combining like terms:
\[ \frac{dy}{dx} = (56x^{107} + 700x^{107}) + (112x^{57} + 700x^{57}) + (500x^{49} + 2000x^{49}) - 168 \]
\[ = 756x^{107} + 812x^{57} + 2500x^{49} - 168 \]

### Final Result
Thus, the derivative of the given function is:
\[ \frac{dy}{dx} = 756x^{107} + 812x^{57} + 2500x^{49} - 168 \]

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