If`y=x^(4)+x^(3)+x^(2)+x+1, x=1 , deltax=0.02,` then `deltay~~....`

Options

A. `0.02`
B. `0.002`
C. `0.2`
D. 2

Answer

To solve the problem, we need to find the approximate change in \( y \) (denoted as \( \Delta y \)) when \( x \) changes by a small amount \( \Delta x \). Here, we have the function:

\[ y = x^4 + x^3 + x^2 + x + 1 \]

Given values are \( x = 1 \) and \( \Delta x = 0.02 \).

### Step 1: Find \( \frac{dy}{dx} \)

To find \( \Delta y \), we first need to compute the derivative \( \frac{dy}{dx} \).

\[ \frac{dy}{dx} = \frac{d}{dx}(x^4 + x^3 + x^2 + x + 1) \]

Using the power rule of differentiation:

\[ \frac{dy}{dx} = 4x^3 + 3x^2 + 2x + 1 \]

### Step 2: Evaluate \( \frac{dy}{dx} \) at \( x = 1 \)

Now, we substitute \( x = 1 \) into the derivative:

\[ \frac{dy}{dx} \bigg|_{x=1} = 4(1)^3 + 3(1)^2 + 2(1) + 1 \]

Calculating this:

\[ = 4 + 3 + 2 + 1 = 10 \]

### Step 3: Use the derivative to find \( \Delta y \)

We can approximate \( \Delta y \) using the formula:

\[ \Delta y \approx \frac{dy}{dx} \cdot \Delta x \]

Substituting the values we have:

\[ \Delta y \approx 10 \cdot 0.02 \]

Calculating this gives:

\[ \Delta y \approx 0.2 \]

### Final Answer

Thus, the approximate change in \( y \) when \( x \) changes by \( 0.02 \) is:

\[ \Delta y \approx 0.2 \]