Math Problem Statement

given:y=x2,::find:\frac{dy}{dx}:when:x=4.:Hence:find:the:approximation:value:of::\left(3.998\right)2

Solution

We are given the function:

y=x2y = x^2

Step 1: Find dydx\frac{dy}{dx}

To differentiate y=x2y = x^2 with respect to xx, we apply the power rule for differentiation:

dydx=2x\frac{dy}{dx} = 2x

Step 2: Find dydx\frac{dy}{dx} when x=4x = 4

Now, we substitute x=4x = 4 into the derivative:

dydx=2(4)=8\frac{dy}{dx} = 2(4) = 8

So, the slope of the curve at x=4x = 4 is 8.

Step 3: Approximate yy for x=3.998x = 3.998

Now, to approximate the value of yy when x=3.998x = 3.998, we can use the tangent line approximation at x=4x = 4. The formula for the tangent line at x=4x = 4 is:

y=y0+dydx(xx0)y = y_0 + \frac{dy}{dx}(x - x_0)

Where:

  • x0=4x_0 = 4
  • y0=x02=16y_0 = x_0^2 = 16 (since x0=4x_0 = 4)
  • dydx=8\frac{dy}{dx} = 8

Now, substitute into the tangent line equation:

y=16+8(3.9984)=16+8(0.002)=160.016=15.984y = 16 + 8(3.998 - 4) = 16 + 8(-0.002) = 16 - 0.016 = 15.984

Approximate value of yy when x=3.998x = 3.998 is y15.984y \approx 15.984.


Would you like further details or have any questions? Here are some related questions:

  1. What is the tangent line approximation for other values of xx?
  2. How can you apply this method to functions other than y=x2y = x^2?
  3. What is the difference between tangent line approximation and linearization?
  4. How do you interpret the slope dydx\frac{dy}{dx} geometrically?
  5. How can you refine approximations using higher-order derivatives?

Tip: Tangent line approximation is particularly useful when the change in xx is small, making it easier to estimate function values near a given point.

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Math Problem Analysis

Mathematical Concepts

Differentiation
Tangent Line Approximation
Power Rule

Formulas

dy/dx = 2x
Tangent Line Approximation: y = y0 + (dy/dx)(x - x0)

Theorems

Power Rule for differentiation

Suitable Grade Level

Grades 11-12