Question

attempt 1: 10 attempts remaining. find (\frac{dy}{dx}) if (y = left(-7x^3 + 9x^2 - x
ight)^3). (\frac{dy}{dx} =)

Explanation:

Step1: Apply the chain rule

Let \( u = -7x^3 + 9x^2 - x \), so \( y = u^3 \). The chain rule states that \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \). First, find \( \frac{dy}{du} \):
For \( y = u^3 \), the derivative with respect to \( u \) is \( \frac{dy}{du} = 3u^2 \).

Step2: Find \( \frac{du}{dx} \)

Now, find the derivative of \( u = -7x^3 + 9x^2 - x \) with respect to \( x \). Using the power rule \( \frac{d}{dx}(x^n) = nx^{n - 1} \):
\( \frac{du}{dx} = -7 \cdot 3x^2 + 9 \cdot 2x - 1 = -21x^2 + 18x - 1 \).

Step3: Substitute back into the chain rule

Substitute \( u = -7x^3 + 9x^2 - x \) and \( \frac{dy}{du} = 3u^2 \), \( \frac{du}{dx} = -21x^2 + 18x - 1 \) into the chain rule formula:
\( \frac{dy}{dx} = 3(-7x^3 + 9x^2 - x)^2 \cdot (-21x^2 + 18x - 1) \).

Answer:

\( 3(-7x^3 + 9x^2 - x)^2(-21x^2 + 18x - 1) \)