Question

question number 2. (4.00 points)
find the derivative of $f(x)=\frac{6}{x^2}-2x^3$.
$\circ f(x)=-6x^2 + 12x$
$\circ f(x)=\frac{6}{x^3}-6x^2$
$\circ f(x)=-\frac{12}{x^3}-6x^2$
$\circ f(x)=\frac{12}{x^3}-6x^2$
$\circ f(x)=-\frac{6}{x^3}+6x^2$
$\circ$none of the above.
question number 3. (4.00 points)

Explanation:

Step1: Rewrite the function

Rewrite \( f(x)=\frac{6}{x^2}-2x^3 \) as \( f(x) = 6x^{-2}-2x^3 \) using the negative exponent rule \( \frac{1}{x^n}=x^{-n} \).

Step2: Apply the power rule

The power rule for differentiation is \( \frac{d}{dx}(x^n)=nx^{n - 1} \).

• For the first term \( 6x^{-2} \), the derivative is \( 6\times(-2)x^{-2-1}=-12x^{-3} \). Using the negative exponent rule again, \( -12x^{-3}=-\frac{12}{x^3} \).
• For the second term \( -2x^3 \), the derivative is \( -2\times3x^{3 - 1}=-6x^2 \).

Step3: Combine the derivatives

Combine the derivatives of the two terms: \( f'(x)=-\frac{12}{x^3}-6x^2 \).

Answer:

\( f'(x)=-\frac{12}{x^3}-6x^2 \) (the third option)