Question

find \\(\frac{dy}{du}\\), \\(\frac{du}{dx}\\), and \\(\frac{dy}{dx}\\).
\\(y = u^{55}\\) and \\(u = 2x^3 - 4x^2\\)
\\(\frac{dy}{du} = 55u^{54}\\)
\\(\frac{du}{dx} = 6x^2 - 8x\\)
\\(\frac{dy}{dx} = \square\\)

Explanation:

Step1: Recall the Chain Rule

The chain rule states that $\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$.

Step2: Substitute the known derivatives

We know $\frac{dy}{du} = 55u^{54}$ and $\frac{du}{dx} = 6x^{2}-8x$. Also, $u = 2x^{3}-4x^{2}$. Substitute $u$ into $\frac{dy}{du}$ and then multiply by $\frac{du}{dx}$:
$\frac{dy}{dx}=55(2x^{3} - 4x^{2})^{54}\cdot(6x^{2}-8x)$
We can factor out 2 from $6x^{2}-8x$ to get $2(3x^{2}-4x)$ and from $2x^{3}-4x^{2}$ to get $2x^{2}(x - 2)$, but for the simplest form, we can also factor $6x^{2}-8x = 2x(3x - 4)$ and $2x^{3}-4x^{2}=2x^{2}(x - 2)$, but the main step is using the chain rule product. Alternatively, just multiply the two expressions:
First, note that $6x^{2}-8x = 2x(3x - 4)$ and $u = 2x^{3}-4x^{2}=2x^{2}(x - 2)$, but the key is substituting $u$ into $\frac{dy}{du}$ and multiplying by $\frac{du}{dx}$. So:
$\frac{dy}{dx}=55(2x^{3}-4x^{2})^{54}\times(6x^{2}-8x)$
We can also factor $6x^{2}-8x = 2(3x^{2}-4x)$ and $2x^{3}-4x^{2}=2x^{2}(x - 2)$, but the standard form after applying chain rule is the product of the two derivatives with $u$ replaced by its expression in $x$.

Answer:

$55(2x^{3}-4x^{2})^{54}(6x^{2}-8x)$ (or factored further as $110x(3x - 4)(2x^{3}-4x^{2})^{54}$ or other factored forms, but the first form is also correct)