Question

Slope of the tangent to the curve $y = 9x^{2} + 7x^{4} + 5$ at the point $x = 1$ is

• A. 28
• B. 16
• C. 46
• D. 1/46

Hint:

Slope = Derivative. Just differentiate and plug in the x-coordinate.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The slope of the tangent to a curve at a given point is equal to the value of the first derivative \( \frac{dy}{dx} \) at that specific point.
Step 2: Key Formula or Approach:
1. Power Rule: \(\frac{d}{dx}(x^n) = nx^{n-1}\).
2. Slope \(m = \left[ \frac{dy}{dx} \right]_{x=1}\).
Step 3: Detailed Explanation:
Given \( y = 9x^2 + 7x^4 + 5 \). Differentiate with respect to \(x\): \[ \frac{dy}{dx} = 9(2x) + 7(4x^3) + 0 \] \[ \frac{dy}{dx} = 18x + 28x^3 \] To find the slope at \(x = 1\), substitute \(x = 1\) into the derivative: \[ m = 18(1) + 28(1)^3 \] \[ m = 18 + 28 = 46 \]
Step 4: Final Answer:
The slope of the tangent at \( x = 1 \) is 46.