Question

attempt 1: 10 attempts remaining.
find \\( \frac{dy}{dx} \\) if \\( y = (-5x^3 + 7x^2 - x)^3 \\).
\\( \frac{dy}{dx} = \\)
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Explanation:

Step1: Identify the outer and inner functions

Let \( u = -5x^3 + 7x^2 - x \), so \( y = u^3 \). We will use the chain rule, which states that \( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx} \).

Step2: Differentiate the outer function with respect to \( u \)

For \( y = u^3 \), the derivative with respect to \( u \) is \( \frac{dy}{du}=3u^2 \) (using the power rule \( \frac{d}{du}(u^n)=nu^{n - 1} \) with \( n = 3 \)).

Step3: Differentiate the inner function with respect to \( x \)

For \( u=-5x^3 + 7x^2 - x \), we differentiate term - by - term:

• The derivative of \( -5x^3 \) with respect to \( x \) is \( - 15x^2 \) (using the power rule \( \frac{d}{dx}(ax^n)=nax^{n - 1} \), here \( a=-5,n = 3 \), so \( 3\times(-5)x^{3 - 1}=-15x^2 \)).
• The derivative of \( 7x^2 \) with respect to \( x \) is \( 14x \) (using the power rule, \( a = 7,n=2 \), so \( 2\times7x^{2 - 1}=14x \)).
• The derivative of \( -x \) with respect to \( x \) is \( - 1 \) (using the power rule, \( a=-1,n = 1 \), so \( 1\times(-1)x^{1 - 1}=-1 \)).

So, \( \frac{du}{dx}=-15x^2 + 14x-1 \).

Step4: Apply the chain rule

Substitute \( \frac{dy}{du}=3u^2 \) and \( \frac{du}{dx}=-15x^2 + 14x - 1 \) into the chain rule formula \( \frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx} \). Since \( u=-5x^3 + 7x^2 - x \), we have:

\( \frac{dy}{dx}=3(-5x^3 + 7x^2 - x)^2(-15x^2 + 14x - 1) \)

Answer:

\( 3(-5x^3 + 7x^2 - x)^2(-15x^2 + 14x - 1) \)