Question

1. if \\( \frac { d y } { d x } = x ^ { 4 } - 2 x ^ { 3 } + 3 x - 1 \\), then \\( \frac { d ^ { 3 } y } { d x ^ { 3 } } \\) evaluated at \\( x = 2 \\) is

Explanation:

Step1: Find the first derivative of \( \frac{dy}{dx} \) (which is \( \frac{d^2y}{dx^2} \))

Given \( \frac{dy}{dx} = x^4 - 2x^3 + 3x - 1 \). Using the power rule \( \frac{d}{dx}(x^n)=nx^{n - 1} \), we have:
\( \frac{d^2y}{dx^2}=\frac{d}{dx}(x^4 - 2x^3 + 3x - 1)=4x^3-6x^2 + 3 \)

Step2: Find the second derivative of \( \frac{dy}{dx} \) (which is \( \frac{d^3y}{dx^3} \))

Differentiate \( \frac{d^2y}{dx^2}=4x^3-6x^2 + 3 \) with respect to \( x \):
\( \frac{d^3y}{dx^3}=\frac{d}{dx}(4x^3-6x^2 + 3)=12x^2-12x \)

Step3: Evaluate \( \frac{d^3y}{dx^3} \) at \( x = 2 \)

Substitute \( x = 2 \) into \( \frac{d^3y}{dx^3}=12x^2-12x \):
\( 12\times(2)^2-12\times(2)=12\times4 - 24=48 - 24 = 24 \)

Answer:

\( 24 \)