Question

question given y = f(u) and u = g(x), find \\(\frac{dy}{dx}\\) by using leibnizs notation for the chain - rule: \\(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\\). y = 5u^4 + 2, u = 3x^3. provide your answer below: \\(\frac{dy}{dx}=\square\\)

Explanation:

Step1: Find $\frac{dy}{du}$

Differentiate $y = 5u^{4}+2$ with respect to $u$. Using the power - rule $\frac{d}{du}(au^{n})=nau^{n - 1}$, we have $\frac{dy}{du}=\frac{d}{du}(5u^{4}+2)=20u^{3}$.

Step2: Find $\frac{du}{dx}$

Differentiate $u = 3x^{3}$ with respect to $x$. Using the power - rule $\frac{d}{dx}(ax^{n})=nax^{n - 1}$, we get $\frac{du}{dx}=\frac{d}{dx}(3x^{3}) = 9x^{2}$.

Step3: Apply the chain - rule

By the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}=20u^{3}$ and $\frac{du}{dx}=9x^{2}$ into the chain - rule formula. Since $u = 3x^{3}$, we have $\frac{dy}{dx}=20(3x^{3})^{3}\cdot9x^{2}$.
Simplify the expression:
\[

$$\begin{align*} \frac{dy}{dx}&=20\times27x^{9}\times9x^{2}\\ &=20\times27\times9x^{9 + 2}\\ &=4860x^{11} \end{align*}$$

\]

Answer:

$4860x^{11}$