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. \(\frac{1}{46}\)

Hint:

The slope of tangent at \(x=a\) is found by calculating \(\left.\frac{dy}{dx}\right|_{x=a}\).

Answer 1

The Correct Option is C

Concept: The slope of the tangent to a curve \(y=f(x)\) at a point is given by: \[ \frac{dy}{dx} \]

Step 1: Given curve: \[ y=9x^2+7x^4+5 \]

Step 2: Differentiate with respect to \(x\). \[ \frac{dy}{dx}=18x+28x^3 \]

Step 3: Put \(x=1\). \[ \left.\frac{dy}{dx}\right|_{x=1}=18(1)+28(1)^3 \] \[ =18+28 \] \[ =46 \] Therefore, slope of the tangent is: \[ \boxed{46} \]