Question

this question is designed to be answered without a calculator. if $y = 8x^2 - 4x^3$, then $y =$ $\circ$ $-24$. $\circ$ $-8$. $\circ$ $0$. $\circ$ $16$.

Explanation:

Step1: Find the first derivative \( y' \)

Using the power rule \( \frac{d}{dx}(x^n) = nx^{n - 1} \), for \( y = 8x^2 - 4x^3 \), we have:
\( y' = \frac{d}{dx}(8x^2) - \frac{d}{dx}(4x^3) = 8\times2x^{2 - 1}-4\times3x^{3 - 1}=16x - 12x^2 \)

Step2: Find the second derivative \( y'' \)

Differentiate \( y' = 16x - 12x^2 \) with respect to \( x \):
\( y'' = \frac{d}{dx}(16x)-\frac{d}{dx}(12x^2)=16\times1x^{1 - 1}-12\times2x^{2 - 1}=16 - 24x \)

Step3: Find the third derivative \( y''' \)

Differentiate \( y'' = 16 - 24x \) with respect to \( x \):
\( y''' = \frac{d}{dx}(16)-\frac{d}{dx}(24x)=0 - 24\times1x^{1 - 1}=-24 \)

Answer:

\(-24\)